Lena Bell

2022-07-09

If $y=\mathrm{ln}(\sqrt{16-{x}^{2}})$, then what is $\frac{dy}{dx}$?

Keegan Barry

Beginner2022-07-10Added 18 answers

This is a chain differentiation

$y=\mathrm{ln}(\sqrt{u}(x))$

$y=\frac{1}{2}\ast \mathrm{ln}(u(x))$

$\frac{dy}{dx}=\frac{1}{2}\ast \frac{1}{u(x)}\ast {u}^{\prime}(x)$

$y=\mathrm{ln}(\sqrt{16-{x}^{2}})$

$y=\frac{1}{2}\ast \mathrm{ln}(16-{x}^{2})$

Therefore,

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

$\frac{dy}{dx}=-\frac{x}{16-{x}^{2}}$

$y=\mathrm{ln}(\sqrt{u}(x))$

$y=\frac{1}{2}\ast \mathrm{ln}(u(x))$

$\frac{dy}{dx}=\frac{1}{2}\ast \frac{1}{u(x)}\ast {u}^{\prime}(x)$

$y=\mathrm{ln}(\sqrt{16-{x}^{2}})$

$y=\frac{1}{2}\ast \mathrm{ln}(16-{x}^{2})$

Therefore,

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

$\frac{dy}{dx}=-\frac{x}{16-{x}^{2}}$

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