Find the derivative of cos ⁡ ( 1 2 π e 6 + 4 x ).

Combine fractions.ddx[cos(πe6+4x2)]Differentiate using the chain rule, which states that&nbsp;ddx[f(g(x))]&nbsp;is&nbsp;f′(g(x))g′(x)&nbsp;where&nbsp;f(x)=cos(x)&nbsp;and&nbsp;g(x)=πe6+4x2.-sin(πe6+4x2)ddx[πe6+4x2]Differentiate using the Constant Multiple Rule.-πsin(πe6+4x2)2ddx[e6+4x]Differentiate using the chain rule, which states that&nbsp;ddx[f(g(x))]&nbsp;is&nbsp;f′(g(x))g′(x)&nbsp;where&nbsp;f(x)=ex&nbsp;and&nbsp;g(x)=6+4x.-πsin(πe6+4x2)2(e6+4xddx[6+4x])Differentiate.-2πe6+4xsin(πe6+4x2)&nbsp;

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datomerki8a5yj

Answered question

2022-05-12

Find the derivative of

\cos (\frac{1}{2} π e^{6 + 4 x})

Answer & Explanation

percolarse2rzd

Beginner2022-05-13Added 17 answers

Combine fractions.

$\frac{d}{d x} [\cos (\frac{π e^{6 + 4 x}}{2})]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = \frac{π e^{6 + 4 x}}{2}$ .

$- \sin (\frac{π e^{6 + 4 x}}{2}) \frac{d}{d x} [\frac{π e^{6 + 4 x}}{2}]$

Differentiate using the Constant Multiple Rule.

$- \frac{π \sin (\frac{π e^{6 + 4 x}}{2})}{2} \frac{d}{d x} [e^{6 + 4 x}]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 6 + 4 x$ .

$- \frac{π \sin (\frac{π e^{6 + 4 x}}{2})}{2} (e^{6 + 4 x} \frac{d}{d x} [6 + 4 x])$

Differentiate.

$- 2 π e^{6 + 4 x} \sin (\frac{π e^{6 + 4 x}}{2})$

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