Question

What is the rate of change $\sqrt{x^2+16}$ with respect to $x^2$ at x = 3 ?

• A. $\frac{1}{5}$
• B. $\frac{1}{10}$
• C. $\frac{1}{20}$
• D. None of the above.

Answer 1

Answer: (B)

$\frac{1}{10}$

Rate of change of both equations is

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

And $\frac{d}{dx}\left(x^2\right)=2x$

Rate of change of $\sqrt{x^2+16}$ w.r.t $x^2$ is

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

At $x=3,$

$\left(\frac{1}{2\sqrt{x^2+16}}\right)=\left(\frac{1}{2\sqrt{3^2+16}}\right)=\frac{1}{2\times 5}=\frac{1}{10}$