question_answer

        Find the slope of the tangent and normal to the curve \[y={{(\sin 2x+\cot x+2)}^{2}}\]At \[x=\frac{\pi }{2}.\]

Answer:

We have, \[y={{(\sin 2x+\cot x+2)}^{2}}\] On- differentiating both sides w.r.t. x, we get \[\frac{dy}{dx}=2(\sin \,2x+\cot \,x+2)\frac{d}{dx}\,(\sin \,\,2x+\cot \,x+2)\]                         [by Chain rule of derivative] \[=2(\sin 2x+\cot x+2)(2\cos 2x-\cos e{{c}^{2}}x)\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{at\,x\,\,=\,\,\frac{\pi }{2}}}=2\left[ \sin \pi -\cot \frac{\pi }{2}+2 \right]\] \[\left[ 2cos\pi -\text{cose}{{\text{c}}^{2}}\frac{\pi }{2} \right]\] \[=2\,[0+0+2][-\,2-1]\] \[=(2\times 2)\times (-\,3)=-12\] \[\therefore \] Slope of tangent \[=-12\] and slope of normal \[=-\frac{1}{{{\left( \frac{dy}{dx} \right)}_{at\,\,\,x\,\,=\,\,\frac{\pi }{2}}}}=-\frac{1}{(-12)}=\frac{1}{12}\]