Question

The rate of change of $\sqrt{x^2+16}$ with respect to $\frac{x}{x-1}$ at x = 3 is ________________.

Answer 1

Let $u\left(x\right)=\sqrt{x^2+16}$ and $v\left(x\right)=\frac{x}{x-1}$.

$u\left(x\right)=\sqrt{x^2+16}$

Differentiating both sides with respect to x, we get

$\frac{du}{dx}=\frac{1}{2\sqrt{x^2+16}}\times 2x=\frac{x}{\sqrt{x^2+16}}$

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

Differentiating both sides with respect to x, we get

$\frac{dv}{dx}=\frac{\left(x-1\right)\times 1-x\times 1}{{\left(x-1\right)}^2}=-\frac{1}{{\left(x-1\right)}^2}$

Now,

Rate of change of u(x) with respect to v(x)

$=\frac{du}{dv}$

$=\frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}$

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

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

∴ Rate of change of u(x) with respect to v(x) at x = 3

$={\left(\frac{du}{dv}\right)}_{x=3}$

$=-\frac{3\times {\left(3-1\right)}^2}{\sqrt{3^2+16}}$

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

Thus, the rate of change of $\sqrt{x^2+16}$ with respect to $\frac{x}{x-1}$ at x = 3 is $-\frac{12}{5}$.


The rate of change of $\sqrt{x^2+16}$ with respect to $\frac{x}{x-1}$ at x = 3 is $\underline{-\frac{12}{5}}$.