What is the derivative and second derivative of $cos (- x)$ $cos(-x)$ ?

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What is the derivative and second derivative of $cos (- x)$ $cos(-x)$ ?

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Answer 1 · 2015-07-02T17:57:28

$\frac{d}{d x} cos (- x) = sin (- x)$ $d/dxcos(-x)=\color(blue)(sin(-x))$ and $\frac{d}{d^{2} x} cos (- x) = - cos (- x)$ $d/(d^2x)cos(-x)=\color(green)(-cos(-x))$

Explanation:

$\frac{d}{d x} cos (- x) = \frac{d}{d x} (- x) \frac{d}{d (- x)} cos (- x)$ $d/dxcos(-x)=d/dx(-x)d/(d(-x))cos(-x)$
..... used the chain rule
$\frac{d}{d x} cos (- x) = - 1 \cdot - sin (- x) = sin (- x)$ $d/dxcos(-x)=-1*-sin(-x)=\color(blue)(sin(-x)$
..... derivative of $cos (x) = - sin (x)$ $cos(x)=-sin(x)$

$\frac{d}{d^{2} x} cos (- x) = \frac{d}{d x} (\frac{d}{d x} cos (- x)) = \frac{d}{d x} (sin (- x))$ $d/(d^2x)cos(-x)=d/dx(d/dxcos(-x))=d/dx(\color(blue)(sin(-x)))$
..... second derivative is derivative of first derivative
$\frac{d}{d^{2} x} cos (- x) = \frac{d}{d x} (- x) \frac{d}{d (- x)} sin (- x)$ $d/(d^2x)cos(-x) = d/dx(-x)d/(d(-x))sin(-x)$
..... used the chain rule again
$\frac{d}{d^{2} x} cos (- x) = - 1 \cdot cos (- x) = - cos (- x)$ $d/(d^2x)cos(-x)=-1*cos(-x) = \color(green)(-cos(-x)$
..... derivative of $sin (x) = cos (x)$ $sin(x)=cos(x)$

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What is the derivative and second derivative of $cos (- x)$ $cos(-x)$ ?

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