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

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Answer 1 · 2016-11-12T11:34:03

$d/dxsqrt(16-x^2)=-1/sqrt(16-x^2)$

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

therefore...
$d/dxsqrt(16-x^2)$

apply chain rule
$1/2(16-x^2)^(-1/2)\timesd/dx(16-x^2)$

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

simplify:
$-1/sqrt(16-x^2)$

How do you differentiate sqrt(16-x^2)sqrt(16-x^2)?