The rate of change
Question:

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

Solution:

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

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

Differentiating both sides with respect to $x$, we get

$\frac{d u}{d x}=\frac{1}{2 \sqrt{x^{2}+16}} \times 2 x=\frac{x}{\sqrt{x^{2}+16}}$

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

Differentiating both sides with respect to $x$, we get

$\frac{d v}{d x}=\frac{(x-1) \times 1-x \times 1}{(x-1)^{2}}=-\frac{1}{(x-1)^{2}}$

Now,

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

$=\frac{d u}{d v}$

$=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

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

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

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

$=\left(\frac{d u}{d v}\right)_{x=3}$

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

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

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

$-\frac{12}{5}$

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