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Please solve rd sharma class 12 chapter 12 derivatives as a rate measure exercise fill in the blanks question 1 maths textbook solution

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Answer: $\frac{-12}{5}$

Hint: Here, we use the basic concept of $\frac{dy}{dx}$ and $\frac{dz}{dx}$

Given: $\sqrt{x^{2}+16}$ with respect to $\left [ \frac{x}{x-1} \right ]$ at $x=3$

Solution: Let $y=\sqrt{x^{2}+16}$

$z=\sqrt{\frac{x}{\left ( x-1 \right )}}$

$\frac{dy}{dx}=x\left ( x^{2}+16 \right )^{\frac{-1}{2}}$

$\frac{dz}{dx}=\left [ \frac{\left \{ \left ( x-1 \right ) -x\right \}}{\left \{ \left ( x-1 \right )^{2} \right \}} \right ]$

$\therefore \frac{dy}{dx}=x\left ( x^{2}+16 \right )^{\frac{-1}{2}}$ and $\left ( \frac{dz}{dx} \right )=\frac{-1}{\left ( x-1 \right )^{2}}$

Hence,

$\frac{dy}{dx}=\left [ \frac{\left ( \frac{dy}{dx} \right )}{\left ( \frac{dz}{dx} \right )} \right ]$

$\therefore \frac{dy}{dx}=\frac{\left [ x\left ( x^{2}+16 \right ) \right ]^{\frac{-1}{2}}}{-1/(x-1)^{2}}$

$\frac{dy}{dz}_{x=3}=\left [ 3\times \frac{1}{5} /\frac{-1}{4}\right ]=\left ( \frac{3}{-5} \right )\times 4$

$=\frac{-12}{5}$

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