A) 2
B) \[\frac{11}{5}\]
C) \[-\frac{12}{5}\]
D) \[-3\]
Correct Answer: C
Solution :
Let \[y=\sqrt{{{x}^{2}}+16}\] and \[z=\frac{x}{x-1}\] Þ\[\frac{dy}{dx}=\frac{1}{2}{{({{x}^{2}}+16)}^{-1/2}}(2x)\]&\[\frac{dz}{dx}=\frac{x-1-x}{{{(x-1)}^{2}}}=\frac{-1}{{{(x-1)}^{2}}}\] \ \[\frac{dy}{dz}=\frac{-x}{\sqrt{{{x}^{2}}+16}}\,\frac{1}{1/{{(x-1)}^{2}}}\] \[{{\left( \frac{dy}{dz} \right)}_{x=3}}=\frac{-3{{(2)}^{2}}}{5}=\frac{-12}{5}\].
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