How do you differentiate $\sqrt{16 - x^{2}}$ $sqrt(16-x^2)$ ?

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How do you differentiate $\sqrt{16 - x^{2}}$ $sqrt(16-x^2)$ ?

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Answer 1 · 2016-11-12T11:34:03

$\frac{d}{d x} \sqrt{16 - x^{2}} = - \frac{1}{\sqrt{16 - x^{2}}}$ $d/dxsqrt(16-x^2)=-1/sqrt(16-x^2)$

Explanation:

diff'n of $\sqrt{x}$ $sqrt(x)$ is $\frac{d}{d x} x^{\frac{1}{2}}$ $d/dxx^(1/2)$
apply $n - 1$ $n-1$ rule: $\frac{1}{2} x^{\frac{1}{2} - 1} \to \frac{1}{2} x^{- \frac{1}{2}}$ $1/2x^(1/2-1)\rightarrow1/2x^(-1/2)$

therefore...
$\frac{d}{d x} \sqrt{16 - x^{2}}$ $d/dxsqrt(16-x^2)$

apply chain rule
$\frac{1}{2} {(16 - x^{2})}^{- \frac{1}{2}} \times \frac{d}{d x} (16 - x^{2})$ $1/2(16-x^2)^(-1/2)\timesd/dx(16-x^2)$

differentiate
$(\frac{1}{2 \times {(16 - x^{2})}^{\frac{1}{2}}}) \times - 2 x$ $(1/(2\times(16-x^2)^(1/2)))\times-2x$

simplify:
$- \frac{1}{\sqrt{16 - x^{2}}}$ $-1/sqrt(16-x^2)$

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How do you differentiate $\sqrt{16 - x^{2}}$ $sqrt(16-x^2)$ ?

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How do you differentiate $\sqrt{16 - x^{2}}$ $sqrt(16-x^2)$ ?