-
question_answer1)
If \[x=a(t-\sin t)\]and \[y=a(1-\cos t),\]then \[\frac{dy}{dx}=\] [AISSE 1984; Roorkee 1974; SCRA 1996; Karnataka CET 2003]
A)
\[\tan \left( \frac{t}{2} \right)\] done
clear
B)
\[-\tan \left( \frac{t}{2} \right)\] done
clear
C)
\[\cot \left( \frac{t}{2} \right)\] done
clear
D)
\[-\cot \left( \frac{t}{2} \right)\] done
clear
View Solution play_arrow
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question_answer2)
If \[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\]and \[y=\frac{2at}{1+{{t}^{2}}}\], then \[\frac{dy}{dx}=\]
A)
\[\frac{a(1-{{t}^{2}})}{2t}\] done
clear
B)
\[\frac{a({{t}^{2}}-1)}{2t}\] done
clear
C)
\[\frac{a({{t}^{2}}+1)}{2t}\] done
clear
D)
\[\frac{a({{t}^{2}}-1)}{t}\] done
clear
View Solution play_arrow
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question_answer3)
If \[x=a\text{ }\left( \cos t+\log \tan \frac{t}{2} \right)\,,y=a\sin t,\]then \[\frac{dy}{dx}=\] [RPET 1997; MP PET 2001]
A)
\[\tan t\] done
clear
B)
\[-\tan t\] done
clear
C)
\[\cot t\] done
clear
D)
\[-\cot t\] done
clear
View Solution play_arrow
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question_answer4)
If \[\tan y=\frac{2t}{1-{{t}^{2}}}\]and \[\sin x=\frac{2t}{1+{{t}^{2}}},\]then \[\frac{dy}{dx}=\]
A)
\[\frac{2}{1+{{t}^{2}}}\] done
clear
B)
\[\frac{1}{1+{{t}^{2}}}\] done
clear
C)
1 done
clear
D)
2 done
clear
View Solution play_arrow
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question_answer5)
If \[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\]and \[y=\frac{2t}{1+{{t}^{2}}}\], then \[\frac{dy}{dx}=\] [Karnataka CET 2000; Pb. CET 2002]
A)
\[\frac{-y}{x}\] done
clear
B)
\[\frac{y}{x}\] done
clear
C)
\[\frac{-x}{y}\] done
clear
D)
\[\frac{x}{y}\] done
clear
View Solution play_arrow
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question_answer6)
If \[x=a{{t}^{2}},y=2at\], then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\] [Karnataka CET 1993]
A)
\[-\frac{1}{{{t}^{2}}}\] done
clear
B)
\[\frac{1}{2a{{t}^{3}}}\] done
clear
C)
\[-\frac{1}{{{t}^{3}}}\] done
clear
D)
\[-\frac{1}{2a{{t}^{3}}}\] done
clear
View Solution play_arrow
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question_answer7)
If \[\cos (x+y)=y\sin x,\]then \[\frac{dy}{dx}=\] [AI CBSE 1979]
A)
\[-\frac{\sin (x+y)+y\cos x}{\sin x+\sin x+y)}\] done
clear
B)
\[\frac{\sin (x+y)+y\cos x}{\sin x+\sin (x+y)}\] done
clear
C)
\[-\frac{\sin (x+y)+y\cos x}{\sin x+\sin x+y)}\] done
clear
D)
d) None of these done
clear
View Solution play_arrow
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question_answer8)
If \[y=\frac{1}{4}{{u}^{4}},u=\frac{2}{3}{{x}^{3}}+5\], then \[\frac{dy}{dx}=\] [DSSE 1979]
A)
\[\frac{1}{27}{{x}^{2}}{{(2{{x}^{3}}+15)}^{3}}\] done
clear
B)
\[\frac{2}{27}x{{(2{{x}^{3}}+5)}^{3}}\] done
clear
C)
\[\frac{2}{27}{{x}^{2}}{{(2{{x}^{3}}+15)}^{3}}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer9)
\[x\sqrt{1+y}+y\sqrt{1+x}=0\], then \[\frac{dy}{dx}=\] [RPET 1989, 96]
A)
\[1+x\] done
clear
B)
\[{{(1+x)}^{-2}}\] done
clear
C)
\[-{{(1+x)}^{-1}}\] done
clear
D)
\[-{{(1+x)}^{-2}}\] done
clear
View Solution play_arrow
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question_answer10)
If \[x=2\cos t-\cos 2t\],\[y=2\sin t-\sin 2t\], then at \[t=\frac{\pi }{4},\frac{dy}{dx}=\]
A)
\[\sqrt{2}+1\] done
clear
B)
\[\sqrt{2+1}\] done
clear
C)
\[\frac{\sqrt{2+1}}{2}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer11)
If \[\sin y=x\sin (a+y),\]then \[\frac{dy}{dx}=\] [Karnataka CET 2000; UPSEAT 2001; Pb. CET 2001; Kerala (Engg.) 2005]
A)
\[\frac{{{\sin }^{2}}(a+y)}{\sin (a+2y)}\] done
clear
B)
\[\frac{{{\sin }^{2}}(a+y)}{\sin (a+2y)}\] done
clear
C)
\[\frac{{{\sin }^{2}}(a+y)}{\sin a}\] done
clear
D)
\[\frac{{{\sin }^{2}}(a+y)}{\cos a}\] done
clear
View Solution play_arrow
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question_answer12)
If \[\tan (x+y)+\tan (x-y)=1,\]then \[\frac{dy}{dx}=\] [DSSE 1979]
A)
\[\frac{{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)}{{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)}\] done
clear
B)
\[\frac{{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)}{{{\sec }^{2}}(x-y)-{{\sec }^{2}}(x+y)}\] done
clear
C)
\[\frac{{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)}{{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer13)
If \[y\sec x+\tan x+{{x}^{2}}y=0\], then \[\frac{dy}{dx}\]= [DSSE 1981; CBSE 1981]
A)
\[\frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{{{x}^{2}}+\sec x}\] done
clear
B)
\[-\frac{2xy+{{\sec }^{2}}x+\sec x\tan x}{{{x}^{2}}+\sec x}\] done
clear
C)
\[-\frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{{{x}^{2}}+\sec x}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer14)
If then \[\frac{dy}{dx}=\] [DSSE 1980; CBSE 1980]
A)
\[\frac{y[2xy-{{y}^{2}}\cos (xy)-1]}{x{{y}^{2}}\cos (xy)+{{y}^{2}}-x}\] done
clear
B)
\[\frac{[2xy-{{y}^{2}}\cos (xy)-1]}{x{{y}^{2}}\cos (xy)+{{y}^{2}}-x}\] done
clear
C)
\[-\frac{y[2xy-{{y}^{2}}\cos (xy)-1]}{x{{y}^{2}}\cos (xy)+{{y}^{2}}-x}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer15)
If \[{{\sin }^{2}}x+2\cos y+xy=0\], then \[\frac{dy}{dx}=\] [AI CBSE 1980]
A)
\[\frac{y+2\sin x}{2\sin y+x}\] done
clear
B)
\[\frac{y+\sin 2x}{2\sin y-x}\] done
clear
C)
\[\frac{y+2\sin x}{\sin y+x}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer16)
If \[{{x}^{3}}+8xy+{{y}^{3}}=64\],then \[\frac{dy}{dx}=\] [AI CBSE 1979]
A)
\[-\frac{3{{x}^{2}}+8y}{8x+3{{y}^{2}}}\] done
clear
B)
\[\frac{3{{x}^{2}}+8y}{8x+3{{y}^{2}}}\] done
clear
C)
\[\frac{3x+8{{y}^{2}}}{8{{x}^{2}}+3y}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer17)
If \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\], then \[\frac{dy}{dx}=\]
A)
\[-\frac{ax+hy+g}{hx-by+f}\] done
clear
B)
\[\frac{ax+hy+g}{hx-by+f}\] done
clear
C)
\[\frac{ax-hy-g}{hx-by-f}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer18)
If \[y=f\left( \frac{5x+1}{10{{x}^{2}}-3} \right)\]and \[f'(x)=\cos x\], then \[\frac{dy}{dx}=\] [MP PET 1987]
A)
\[\cos \,\left( \frac{5x+1}{10{{x}^{2}}-3} \right)\,\frac{dy}{dx}\,\left( \frac{5x+1}{10{{x}^{2}}-3} \right)\] done
clear
B)
\[\frac{5x+1}{10{{x}^{2}}-3}\,\,\cos \,\left( \frac{5x+1}{10{{x}^{2}}-3} \right)\] done
clear
C)
\[\cos \,\left( \frac{5x+1}{10{{x}^{2}}-3} \right)\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer19)
If\[f(x)=\frac{1}{1-x}\], then the derivative of the composite function \[f[f\{f(x)\}]\] is equal to [Orissa JEE 2003]
A)
0 done
clear
B)
\[\frac{1}{2}\] done
clear
C)
1 done
clear
D)
2 done
clear
View Solution play_arrow
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question_answer20)
Let g (x) be the inverse of an invertible function \[f(x)\] which is differentiable at x = c, then \[g'(f(c))\]equals
A)
\[f'(c)\] done
clear
B)
\[\frac{1}{f'(c)}\] done
clear
C)
\[f(c)\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer21)
Let \[g(x)\] be the inverse of the function \[f(x)\] and \[f'(x)=\frac{1}{1+{{x}^{3}}}\]. Then \[{g}'(x)\] is equal to [Kurukshetra CEE 1996]
A)
\[\frac{1}{1+{{(g(x))}^{3}}}\] done
clear
B)
\[\frac{1}{1+{{(f(x))}^{3}}}\] done
clear
C)
\[1+{{(g(x))}^{3}}\] done
clear
D)
\[1+{{(f(x))}^{3}}\] done
clear
View Solution play_arrow
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question_answer22)
Let f and g be differentiable functions satisfying \[{g}'(a)=2,\] \[g(a)=b\] and \[fog=I\](identity function). Then \[f'(b)\] is equal to
A)
\[\frac{1}{2}\] done
clear
B)
2 done
clear
C)
\[\frac{2}{3}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer23)
The differential coefficient of \[f[\log (x)]\] when \[f(x)=\log x\]is [Kurukshetra CEE 1998; DCE 2000]
A)
\[x\log x\] done
clear
B)
\[\frac{x}{\log x}\] done
clear
C)
\[\frac{1}{x\log x}\] done
clear
D)
\[\frac{\log x}{x}\] done
clear
View Solution play_arrow
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question_answer24)
The derivative of \[F[f\{\varphi (x)\}]\] is [AMU 2001]
A)
\[{F}'\,[f\,\{\varphi \,(x)\}]\] done
clear
B)
\[F\,[f\,\{\varphi \,(x)\}\,]\,{f}'\{\varphi (x)\}\] done
clear
C)
\[{F}'[f\,\{\varphi \,(x)\}]\,{f}'\{\varphi (x)\}\] done
clear
D)
\[{F}'\,[f\,\{\varphi \,(x)\}]\,{f}'\{\varphi (x)\}\,{\varphi }'\,(x)\] done
clear
View Solution play_arrow
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question_answer25)
Let \[f(x)={{e}^{x}}\], \[g(x)={{\sin }^{-1}}x\] and \[h(x)=f(g(x)),\] then \[h'(x)/h(x)=\] [EAMCET 2002]
A)
\[{{e}^{{{\sin }^{-1}}x}}\] done
clear
B)
\[1/\sqrt{1-{{x}^{2}}}\] done
clear
C)
\[{{\sin }^{-1}}x\] done
clear
D)
\[1/\,(1-{{x}^{2}})\] done
clear
View Solution play_arrow
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question_answer26)
If \[{{x}^{2}}+{{y}^{2}}=t-\frac{1}{t},\]\[{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\], then \[{{x}^{3}}y\frac{dy}{dx}=\]
A)
1 done
clear
B)
2 done
clear
C)
3 done
clear
D)
4 done
clear
View Solution play_arrow
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question_answer27)
If \[x=a\sin 2\theta (1+\cos 2\theta ),y=b\cos 2\theta (1-\cos 2\theta )\], then \[\frac{dy}{dx}=\] [Kurukshetra CEE 1998]
A)
\[\frac{b\tan \theta }{a}\] done
clear
B)
\[\frac{a\tan \theta }{b}\] done
clear
C)
\[\frac{a}{b\tan \theta }\] done
clear
D)
\[\frac{b}{a\tan \theta }\] done
clear
View Solution play_arrow
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question_answer28)
If \[\sin y=x\cos (a+y),\] then \[\frac{dy}{dx}=\]
A)
\[\frac{{{\cos }^{2}}(a+y)}{\cos a}\] done
clear
B)
\[\frac{\cos (a+y)}{{{\cos }^{2}}a}\] done
clear
C)
\[\frac{{{\sin }^{2}}(a+y)}{\sin a}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer29)
If \[x=\frac{3at}{1+{{t}^{3}}},y=\frac{3a{{t}^{2}}}{1+{{t}^{3}}},\]then \[\frac{dy}{dx}\]=
A)
\[\frac{t(2+{{t}^{3}})}{1-2{{t}^{3}}}\] done
clear
B)
\[\frac{t(2-{{t}^{3}})}{1-2{{t}^{3}}}\] done
clear
C)
\[\frac{t(2+{{t}^{3}})}{1+2{{t}^{3}}}\] done
clear
D)
\[\frac{t(2-{{t}^{3}})}{1+2{{t}^{3}}}\] done
clear
View Solution play_arrow
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question_answer30)
If \[x=t+\frac{1}{t},y=t-\frac{1}{t},\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is equal to
A)
\[-4t{{({{t}^{2}}-1)}^{-2}}\] done
clear
B)
\[-4{{t}^{3}}{{({{t}^{2}}-1)}^{-3}}\] done
clear
C)
\[({{t}^{2}}+1){{({{t}^{2}}-1)}^{-1}}\] done
clear
D)
\[-4{{t}^{2}}{{({{t}^{2}}-1)}^{-2}}\] done
clear
View Solution play_arrow
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question_answer31)
If \[x={{t}^{2}}\], \[y={{t}^{3}}\], then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\] = [EAMCET 1994]
A)
\[\frac{3}{2}\] done
clear
B)
\[\frac{3}{(4t)}\] done
clear
C)
\[\frac{3}{2(t)}\] done
clear
D)
\[\frac{3t}{2}\] done
clear
View Solution play_arrow
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question_answer32)
If \[x=a\sin \theta \] and \[y=b\]\[\cos \theta ,\] then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\] is [UPSEAT 2002]
A)
\[\frac{a}{{{b}^{2}}}{{\sec }^{2}}\theta \] done
clear
B)
\[\frac{-b}{a}{{\sec }^{2}}\theta \] done
clear
C)
\[\frac{-b}{{{a}^{2}}}{{\sec }^{3}}\theta \] done
clear
D)
\[\frac{-b}{{{a}^{2}}}{{\sec }^{3}}\theta \] done
clear
View Solution play_arrow
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question_answer33)
Let \[y={{t}^{10}}+1\]and \[x={{t}^{8}}+1,\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is [UPSEAT 2004]
A)
\[\frac{5}{2}t\] done
clear
B)
\[20{{t}^{8}}\] done
clear
C)
\[\frac{5}{16{{t}^{6}}}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer34)
If \[3\sin (xy)+4\cos (xy)=5\], then \[\frac{dy}{dx}=\] [EAMCET 1994]
A)
\[-\frac{y}{x}\] done
clear
B)
\[\frac{3\sin (xy)+4\cos (xy)}{3\cos (xy)-4\sin (xy)}\] done
clear
C)
\[\frac{3\cos (xy)+4\sin (xy)}{4\cos (xy)-3\sin (xy)}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer35)
If \[{{x}^{2}}{{e}^{y}}+2xy{{e}^{x}}+13=0\], then dy/dx = [RPET 1987]
A)
\[\frac{2x{{e}^{y-x}}+2y(x+1)}{x(x{{e}^{y-x}}+2)}\] done
clear
B)
\[\frac{2x{{e}^{x-y}}+2y(x+1)}{x(x{{e}^{y-x}}+2)}\] done
clear
C)
\[-\frac{2x{{e}^{y-x}}+2y(x+1)}{x(x{{e}^{y-x}}+2)}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer36)
If \[x=a{{\cos }^{3}}\theta ,y=a{{\sin }^{3}}\theta \], then \[\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}=\] [EAMCET 1992]
A)
\[{{\tan }^{2}}\theta \] done
clear
B)
\[{{\sec }^{2}}\theta \] done
clear
C)
\[\sec \theta \] done
clear
D)
\[|\sec \theta |\] done
clear
View Solution play_arrow
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question_answer37)
If \[{{x}^{3}}+{{y}^{3}}-3axy=0\], then \[\frac{dy}{dx}\] equals [RPET 1996]
A)
\[\frac{ay-{{x}^{2}}}{{{y}^{2}}-ax}\] done
clear
B)
\[\frac{ay-{{x}^{2}}}{ay-{{y}^{2}}}\] done
clear
C)
\[\frac{{{x}^{2}}+ay}{{{y}^{2}}+ax}\] done
clear
D)
\[\frac{{{x}^{2}}+ay}{ax-{{y}^{2}}}\] done
clear
View Solution play_arrow
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question_answer38)
If \[x=a(t+\sin t)\]and \[y=a(1-\cos t)\], then \[\frac{dy}{dx}\] equals [RPET 1996; MP PET 2002]
A)
\[\tan (t/2)\] done
clear
B)
\[\cot (t/2)\] done
clear
C)
\[\tan 2t\] done
clear
D)
\[\tan t\] done
clear
View Solution play_arrow
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question_answer39)
If \[x=\frac{2\,t}{1+{{t}^{2}}},\,\,y=\frac{1-{{t}^{2}}}{1+{{t}^{2}}},\]then \[\frac{d\,y}{d\,x}\] equals [RPET 1999]
A)
\[\frac{2\,t}{{{t}^{2}}+1}\] done
clear
B)
\[\frac{2\,t}{{{t}^{2}}-1}\] done
clear
C)
\[\frac{2\,t}{1-{{t}^{2}}}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer40)
If sin(x+y)=log(x+y), then \[\frac{dy}{dx}\]= [Karnataka CET 1993; RPET 1989, 92; Roorkee 2000]
A)
2 done
clear
B)
? 2 done
clear
C)
1 done
clear
D)
?1 done
clear
View Solution play_arrow
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question_answer41)
If \[\cos x=\frac{1}{\sqrt{1+{{t}^{2}}}}\]and \[\sin y=\frac{t}{\sqrt{1+{{t}^{2}}}}\], then \[\frac{dy}{dx}=\] [MP PET 1994]
A)
?1 done
clear
B)
\[\frac{1-t}{1+{{t}^{2}}}\] done
clear
C)
\[\frac{1}{1+{{t}^{2}}}\] done
clear
D)
1 done
clear
View Solution play_arrow
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question_answer42)
If\[x=a(\cos \theta +\theta \sin \theta )\], \[y=a(\sin \theta -\theta \cos \theta ),\text{ }\]then \[\frac{dy}{dx}=\] [DCE 1999]
A)
\[\cos \theta \] done
clear
B)
\[\tan \theta \] done
clear
C)
\[\sec \theta \] done
clear
D)
cosecq done
clear
View Solution play_arrow
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question_answer43)
If \[x=a{{\cos }^{4}}\theta ,y=a{{\sin }^{4}}\theta ,\] then \[\frac{dy}{dx}\], at \[\theta =\frac{3\pi }{4}\], is [Kerala (Engg.) 2002]
A)
?1 done
clear
B)
1 done
clear
C)
\[-{{a}^{2}}\] done
clear
D)
\[{{a}^{2}}\] done
clear
View Solution play_arrow
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question_answer44)
If \[x={{\sin }^{-1}}(3t-4{{t}^{3}})\] and \[y={{\cos }^{-1}}\,\,\sqrt{(1-{{t}^{2}})}\], then \[\frac{dy}{dx}\] is equal to [Kerala (Engg.) 2002]
A)
½ done
clear
B)
2/5 done
clear
C)
3/2 done
clear
D)
1/3 done
clear
View Solution play_arrow
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question_answer45)
If \[x=a\left( t-\frac{1}{t} \right)\,,y=a\] \[\left( t+\frac{1}{t} \right)\]then \[\frac{dy}{dx}=\] [Karnataka CET 2004]
A)
\[\frac{y}{x}\] done
clear
B)
\[\frac{-y}{x}\] done
clear
C)
\[\frac{x}{y}\] done
clear
D)
\[\frac{-x}{y}\] done
clear
View Solution play_arrow
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question_answer46)
If \[x=\sin t\cos 2t\] and \[y=\cos t\sin 2t\], then at \[t=\frac{\pi }{4},\] the value of \[\frac{dy}{dx}\] is equal to [Pb. CET 2000]
A)
?2 done
clear
B)
2 done
clear
C)
\[\frac{1}{2}\] done
clear
D)
\[-\frac{1}{2}\] done
clear
View Solution play_arrow
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question_answer47)
If \[\ln \,(x+y)=2xy,\]then \[y'(0)\]= [IIT Screening 2004]
A)
1 done
clear
B)
?1 done
clear
C)
2 done
clear
D)
0 done
clear
View Solution play_arrow
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question_answer48)
If \[y={{x}^{x}}\], then \[\frac{dy}{dx}=\] [AISSE 1984; DSSE 1982; MNR 1979; SCRA 1996; RPET 1996; Kerala (Engg.) 2002]
A)
\[{{x}^{x}}\log ex\] done
clear
B)
\[{{x}^{x}}\left( 1+\frac{1}{x} \right)\] done
clear
C)
\[(1+\log x)\] done
clear
D)
\[{{x}^{x}}\log x\] done
clear
View Solution play_arrow
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question_answer49)
The first derivative of the function \[\left[ {{\cos }^{-1}}\left( \sin \sqrt{\frac{1+x}{2}} \right)+{{x}^{x}} \right]\] with respect to x at x = 1 is [MP PET 1998]
A)
\[\frac{3}{4}\] done
clear
B)
0 done
clear
C)
\[\frac{1}{2}\] done
clear
D)
\[-\frac{1}{2}\] done
clear
View Solution play_arrow
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question_answer50)
If \[y=\sqrt{\frac{1+x}{1-x}},\]then \[\frac{dy}{dx}=\] [AISSE 1981; RPET 1995]
A)
\[\frac{2}{{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}}\] done
clear
B)
\[\frac{1}{{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}}\] done
clear
C)
\[\frac{1}{2{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}}\] done
clear
D)
\[\frac{1}{{{(1+x)}^{3/2}}{{(1-x)}^{1/2}}}\] done
clear
View Solution play_arrow
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question_answer51)
If \[y={{e}^{x+{{e}^{x+{{e}^{x+....\infty }}}}}}\], then \[\frac{dy}{dx}=\] [AISSE 1990; UPSEAT 2002; DCE 2002]
A)
\[\frac{y}{1-y}\] done
clear
B)
\[\frac{1}{1-y}\] done
clear
C)
\[\frac{y}{1+y}\] done
clear
D)
\[\frac{y}{y-1}\] done
clear
View Solution play_arrow
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question_answer52)
If \[{{x}^{y}}={{e}^{x-y}}\], then \[\frac{dy}{dx}=\] [MP PET 1987, 2004; MNR 1984; Roorkee 1954; BIT Ranchi 1991; RPET 2000]
A)
\[\log x.{{[\log (ex)]}^{-2}}\] done
clear
B)
\[\log x.{{[\log (ex)]}^{2}}\] done
clear
C)
\[\log x.{{(\log x)}^{2}}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer53)
\[(x-y){{e}^{x/(x-y)}}=k\] then
A)
\[(y-2x)\frac{dy}{dx}+3x-2y=0\] done
clear
B)
\[y\frac{dy}{dx}+x-2y=0\] done
clear
C)
\[a\text{ }\left( y\frac{dy}{dx}+x-2y \right)=0\] done
clear
D)
None of these done
clear
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question_answer54)
If \[{{2}^{x}}+{{2}^{y}}={{2}^{x+y}}\], then \[\frac{dy}{dx}=\] [MP PET 1995; AMU 2000]
A)
\[{{2}^{x-y}}\frac{{{2}^{y}}-1}{{{2}^{x}}-1}\] done
clear
B)
\[{{2}^{x-y}}\frac{{{2}^{y}}-1}{1-{{2}^{x}}}\] done
clear
C)
\[\frac{{{2}^{x}}+{{2}^{y}}}{{{2}^{x}}-{{2}^{y}}}\] done
clear
D)
None of these done
clear
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question_answer55)
If \[y=\log {{x}^{x}},\]then \[\frac{dy}{dx}=\] [MNR 1978]
A)
\[{{x}^{x}}(1+\log x)\] done
clear
B)
\[\log (ex)\] done
clear
C)
\[\log \left( \frac{e}{x} \right)\] done
clear
D)
None of these done
clear
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-
question_answer56)
If \[{{y}^{x}}+{{x}^{y}}={{a}^{b}}\],then \[\frac{dy}{dx}=\]
A)
\[-\frac{y{{x}^{y-1}}+{{y}^{x}}\log y}{x{{y}^{x-1}}+{{x}^{y}}\log x}\] done
clear
B)
\[\frac{y{{x}^{y-1}}+{{y}^{x}}\log y}{x{{y}^{x-1}}+{{x}^{y}}\log x}\] done
clear
C)
\[-\frac{y{{x}^{y-1}}+{{y}^{x}}}{x{{y}^{x-1}}+{{x}^{y}}l}\] done
clear
D)
\[\frac{y{{x}^{y-1}}+{{y}^{x}}}{x{{y}^{x-1}}+{{x}^{y}}}\] done
clear
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question_answer57)
If \[y=\sqrt{\frac{(x-a)(x-b)}{(x-c)(x-d)}}\], then \[\frac{dy}{dx}=\]
A)
\[\frac{y}{2}\left[ \frac{1}{x-a}+\frac{1}{x-b}-\frac{1}{x-c}-\frac{1}{x-d} \right]\] done
clear
B)
\[y\,\left[ \frac{1}{x-a}+\frac{1}{x-b}-\frac{1}{x-c}-\frac{1}{x-d} \right]\] done
clear
C)
\[\frac{1}{2}\left[ \frac{1}{x-a}+\frac{1}{x-b}-\frac{1}{x-c}-\frac{1}{x-d} \right]\] done
clear
D)
None of these done
clear
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question_answer58)
If \[y={{(1+x)}^{x}},\]then \[\frac{dy}{dx}=\]
A)
\[{{(1+x)}^{x}}\left[ \frac{x}{1+x}+\log ex \right]\] done
clear
B)
\[\frac{x}{1+x}+\log (1+x)\] done
clear
C)
\[{{(1+x)}^{x}}\left[ \frac{x}{1+x}+\log (1+x) \right]\] done
clear
D)
None of these done
clear
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question_answer59)
If \[y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+.....\infty }}}\], then \[\frac{dy}{dx}=\]
A)
\[\frac{x}{2y-1}\] done
clear
B)
\[\frac{x}{2y+1}\] done
clear
C)
\[\frac{1}{x(2y-1)}\] done
clear
D)
\[\frac{1}{x(1-2y)}\] done
clear
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question_answer60)
If \[y={{x}^{\sqrt{x}}},\]then \[\frac{dy}{dx}\]=
A)
\[{{x}^{\sqrt{x}}}\frac{2+\log x}{2\sqrt{x}}\] done
clear
B)
\[{{x}^{\sqrt{x}}}\frac{2+\log x}{\sqrt{x}}\] done
clear
C)
\[\frac{2+\log x}{2\sqrt{x}}\] done
clear
D)
None of these done
clear
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question_answer61)
If \[{{x}^{p}}{{y}^{q}}={{(x+y)}^{p+q}},\]then \[\frac{dy}{dx}=\] [RPET 1999; UPSEAT 2001]
A)
\[\frac{y}{x}\] done
clear
B)
\[-\frac{y}{x}\] done
clear
C)
\[\frac{x}{y}\] done
clear
D)
\[-\frac{x}{y}\] done
clear
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question_answer62)
If \[y={{(\sin x)}^{{{(\sin x)}^{(\sin x)......\infty }}}}\], then \[\frac{dy}{dx}=\]
A)
\[\frac{{{y}^{2}}\cot x}{1-y\log \sin x}\] done
clear
B)
\[\frac{{{y}^{2}}\cot x}{1+y\log \sin x}\] done
clear
C)
\[\frac{y\cot x}{1-y\log \sin x}\] done
clear
D)
\[\frac{y\cot x}{1+y\log \sin x}\] done
clear
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question_answer63)
If \[y={{({{x}^{x}})}^{x}}\], then \[\frac{dy}{dx}\]=
A)
\[{{({{x}^{x}})}^{x}}(1+2\log x)\] done
clear
B)
\[{{({{x}^{x}})}^{x}}(1+\log x)\] done
clear
C)
\[x{{({{x}^{x}})}^{x}}(1+2\log x)\] done
clear
D)
\[x\,{{({{x}^{x}})}^{x}}(1+\log x)\] done
clear
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question_answer64)
The differential equation satisfied by the function \[y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....\infty }}}\], is [MP PET 1998; Pb. CET 2001]
A)
\[(2y-1)\frac{dy}{dx}-\sin x=0\] done
clear
B)
\[(2y-1)\cos x+\frac{dy}{dx}=0\] done
clear
C)
\[(2y-1)\cos x-\frac{dy}{dx}=0\] done
clear
D)
\[(2y-1)\cos x+\frac{dy}{dx}=0\] done
clear
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question_answer65)
If \[y={{\left( 1+\frac{1}{x} \right)}^{x}}\], then \[\frac{dy}{dx}=\] [BIT Ranchi 1992]
A)
\[{{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right)-\frac{1}{1+x} \right]\] done
clear
B)
\[{{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right) \right]\] done
clear
C)
\[{{\left( x+\frac{1}{x} \right)}^{x}}\left[ \log (x-1)-\frac{x}{x+1} \right]\] done
clear
D)
\[{{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right)+\frac{1}{1+x} \right]\] done
clear
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question_answer66)
\[\frac{d}{dx}({{x}^{{{\log }_{e}}x}})=\] [MP PET 1993]
A)
\[2{{x}^{({{\log }_{e}}x-1)}}.{{\log }_{e}}x\] done
clear
B)
\[{{x}^{({{\log }_{e}}x-1)}}\] done
clear
C)
\[\frac{2}{x}{{\log }_{e}}x\] done
clear
D)
\[{{x}^{({{\log }_{e}}x-1)}}.{{\log }_{e}}x\] done
clear
View Solution play_arrow
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question_answer67)
If \[{{x}^{y}}={{y}^{x}},\]then \[\frac{dy}{dx}=\] [DSSE 1996; MP PET 1997]
A)
\[\frac{y(x{{\log }_{e}}y+y)}{x(y{{\log }_{e}}x+x)}\] done
clear
B)
\[\frac{y(x{{\log }_{e}}y-y)}{x(y{{\log }_{e}}x-x)}\] done
clear
C)
\[\frac{x(x{{\log }_{e}}y-y)}{y(y{{\log }_{e}}x-x)}\] done
clear
D)
\[\frac{x(x{{\log }_{e}}y+y)}{y(y{{\log }_{e}}x+x)}\] done
clear
View Solution play_arrow
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question_answer68)
If \[y={{x}^{({{x}^{x}})}}\], then \[\frac{dy}{dx}=\] [AISSE 1989]
A)
\[y[{{x}^{x}}(\log ex).\log x+{{x}^{x}}]\] done
clear
B)
\[y[{{x}^{x}}(\log ex).\log x+x]\] done
clear
C)
\[y[{{x}^{x}}(\log ex).\log x+{{x}^{x-1}}]\] done
clear
D)
\[y[{{x}^{x}}({{\log }_{e}}x).\log x+{{x}^{x-1}}]\] done
clear
View Solution play_arrow
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question_answer69)
If \[y={{x}^{\sin x}},\]then \[\frac{dy}{dx}=\] [DSSE 1983, 84]
A)
\[\frac{x\cos x.\log x+\sin x}{x}.{{x}^{\sin x}}\] done
clear
B)
\[\frac{y[x\cos x.\log x+\cos x]}{x}\] done
clear
C)
\[y[x\sin x.\log x+\cos x]\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer70)
\[\frac{d}{dx}\{{{(\sin x)}^{x}}\}\]= [DSSE 1985, 87; AISSE 1983]
A)
\[\left[ \frac{x\cos x+\sin x\log \sin x}{\sin x} \right]\] done
clear
B)
\[{{(\sin x)}^{x}}\left[ \frac{x\cos x+\sin x\log \sin x}{\sin x} \right]\] done
clear
C)
\[{{(\sin x)}^{x}}\left[ \frac{x\sin x+\sin x\log \sin x}{\sin x} \right]\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer71)
If \[y=\frac{\sqrt{x}{{(2x+3)}^{2}}}{\sqrt{x+1}},\]then \[\frac{dy}{dx}=\] [AISSE 1986]
A)
\[y\text{ }\left[ \frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)} \right]\] done
clear
B)
\[y\text{ }\left[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{2(x+1)} \right]\] done
clear
C)
\[y\text{ }\left[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{x+1} \right]\] done
clear
D)
None of these done
clear
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question_answer72)
\[\frac{d}{dx}\{{{(\sin x)}^{\log x}}\}=\] [DSSE 1984]
A)
\[{{(\sin x)}^{\log x}}\left[ \frac{1}{x}\log \sin x+\cot x \right]\] done
clear
B)
\[{{(\sin x)}^{\log x}}\left[ \frac{1}{x}\log \sin x+\cot x\log x \right]\] done
clear
C)
\[{{(\sin x)}^{\log x}}\left[ \frac{1}{x}\log \sin x+\log x \right]\] done
clear
D)
None of these done
clear
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question_answer73)
If \[y={{(\tan x)}^{\cot x}}\], then \[\frac{dy}{dx}\backslash \]= [AISSE 1985]
A)
\[y\cos \text{e}{{\text{c}}^{2}}x(1-\log \tan x)\] done
clear
B)
\[y\,\text{cos}\text{e}{{\text{c}}^{2}}x(1+\log \tan x)\] done
clear
C)
\[y\cos \text{e}{{\text{c}}^{2}}x(\log \tan x)\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer74)
If \[y={{x}^{2}}+{{x}^{\log x}},\]then \[\frac{dy}{dx}=\] [DSSE 1986]
A)
\[\frac{{{x}^{2}}+\log x.{{x}^{\log x}}}{x}\] done
clear
B)
\[{{x}^{2}}+\log x.{{x}^{\log x}}\] done
clear
C)
\[\frac{2({{x}^{2}}+\log x.{{x}^{\log x}})}{x}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer75)
If \[y={{x}^{2}}+\frac{1}{{{x}^{2}}+\frac{1}{{{x}^{2}}+\frac{1}{{{x}^{2}}+......\infty }}},\]then \[\frac{dy}{dx}=\]
A)
\[\frac{2xy}{2y-{{x}^{2}}}\] done
clear
B)
\[\frac{xy}{y+{{x}^{2}}}\] done
clear
C)
\[\frac{xy}{y-{{x}^{2}}}\] done
clear
D)
\[\frac{2xy}{2+\frac{{{x}^{2}}}{y}}\] done
clear
View Solution play_arrow
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question_answer76)
If \[y={{\sqrt{x}}^{{{\sqrt{x}}^{\sqrt{x}....\infty }}}}\], then \[\frac{dy}{dx}=\]
A)
\[\frac{{{y}^{2}}}{2x-2y\log x}\] done
clear
B)
\[\frac{{{y}^{2}}}{2x+\log x}\] done
clear
C)
\[\frac{{{y}^{2}}}{2x+2y\log x}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer77)
If \[{{x}^{y}}.{{y}^{x}}=1\], then \[\frac{dy}{dx}\]=
A)
\[\frac{y\,(y+x\log y)}{x(y\log x+x)}\] done
clear
B)
\[\frac{y\,(x+y\log x)}{x(y+x\log y)}\] done
clear
C)
\[-\frac{y(y+x\log y)}{x(x+y\log x)}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer78)
\[y={{(\tan x)}^{{{(\tan x)}^{\tan x}}}},\] then at\[x=\frac{\pi }{4}\], the value of \[\frac{dy}{dx}=\] [WB JEE 1990]
A)
0 done
clear
B)
1 done
clear
C)
2 done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer79)
If \[y={{(\sin x)}^{\tan x}}\], then \[\frac{dy}{dx}\]is equal to [IIT 1994; RPET 1996]
A)
\[{{(\sin x)}^{\tan x}}.(1+{{\sec }^{2}}x.\log \sin x)\] done
clear
B)
\[\tan x.{{(\sin x)}^{\tan x-1}}.\cos x\] done
clear
C)
\[{{(\sin x)}^{\tan x}}.{{\sec }^{2}}x.\log \sin x\] done
clear
D)
\[\tan x.{{(\sin x)}^{\tan x-1}}\] done
clear
View Solution play_arrow
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question_answer80)
If \[y={{2}^{1/{{\log }_{x}}4}}\], then x is equal to [Roorkee 1998]
A)
\[\sqrt{y}\] done
clear
B)
\[y\] done
clear
C)
\[{{y}^{2}}\] done
clear
D)
\[{{y}^{4}}\] done
clear
View Solution play_arrow
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question_answer81)
If \[{{2}^{x}}+{{2}^{y}}={{2}^{x+y}},\]then the value of \[\frac{dy}{dx}\] at \[x=y=1\]is [Karnataka CET 2000]
A)
0 done
clear
B)
? 1 done
clear
C)
1 done
clear
D)
2 done
clear
View Solution play_arrow
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question_answer82)
The derivative of \[y={{x}^{\ln x}}\] is [AMU 2000]
A)
\[{{x}^{\ln x}}\ln x\] done
clear
B)
\[{{x}^{\text{ln}\,x-1}}\text{ln}\,x\] done
clear
C)
\[2{{x}^{\ln x-1}}\ln \,x\] done
clear
D)
\[{{x}^{\ln x-2}}\] done
clear
View Solution play_arrow
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question_answer83)
If \[y=\sqrt{x+\sqrt{x+\sqrt{x+........\text{to}}}}\infty \,,\,\text{then}\frac{dy}{dx}=\] [RPET 2002]
A)
\[\frac{x}{2y-1}\] done
clear
B)
\[\frac{2}{2y-1}\] done
clear
C)
\[\frac{-1}{2y-1}\] done
clear
D)
\[\frac{1}{2y-1}\] done
clear
View Solution play_arrow
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question_answer84)
If \[{{x}^{m}}{{y}^{n}}=2{{(x+y)}^{m+n}},\] the value of \[\frac{dy}{dx}\] is [MP PET 2003]
A)
\[x+y\] done
clear
B)
\[x/y\] done
clear
C)
\[y/x\] done
clear
D)
\[x-y\] done
clear
View Solution play_arrow
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question_answer85)
If \[x={{e}^{y+{{e}^{y+....t\text{o}\,\,\infty }}}}\], \[x>0,\] then \[\frac{dy}{dx}\] is [AIEEE 2004]
A)
\[\frac{1+x}{x}\] done
clear
B)
\[\frac{1}{x}\] done
clear
C)
\[\frac{1-x}{x}\] done
clear
D)
\[\frac{x}{1+x}\] done
clear
View Solution play_arrow