# If y = ln &#x2061;<!-- ⁡ --> ( <msqrt> 16 &#x2212;<!-- - --> x

If $y=\mathrm{ln}\left(\sqrt{16-{x}^{2}}\right)$, then what is $\frac{dy}{dx}$?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Keegan Barry
This is a chain differentiation
$y=\mathrm{ln}\left(\sqrt{u}\left(x\right)\right)$
$y=\frac{1}{2}\ast \mathrm{ln}\left(u\left(x\right)\right)$
$\frac{dy}{dx}=\frac{1}{2}\ast \frac{1}{u\left(x\right)}\ast {u}^{\prime }\left(x\right)$
$y=\mathrm{ln}\left(\sqrt{16-{x}^{2}}\right)$
$y=\frac{1}{2}\ast \mathrm{ln}\left(16-{x}^{2}\right)$
Therefore,
$\frac{dy}{dx}=\frac{1}{2}\ast \frac{1}{16-{x}^{2}}\ast \left(-2x\right)$
$\frac{dy}{dx}=-\frac{x}{16-{x}^{2}}$