How do you find the derivative of $({sin}^{2} x) ({cos}^{3} x)$ $(sin^2x)(cos^3x)$ ?

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How do you find the derivative of $({sin}^{2} x) ({cos}^{3} x)$ $(sin^2x)(cos^3x)$ ?

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Answer 1 · 2017-07-13T15:08:00

$\frac{d}{d x} [({cos}^{3} x) ({sin}^{2} x)] = 2 {cos}^{4} x sin x - 3 {cos}^{2} x {sin}^{3} x$ $d/(dx)[(cos^3x)(sin^2x)] = color(blue)(2cos^4xsinx - 3cos^2xsin^3x$

Explanation:

We're asked to find the derivative

$\frac{d}{d x} [({cos}^{3} x) ({sin}^{2} x)]$ $d/(dx) [(cos^3x)(sin^2x)]$

The first step we could do is use the product rule, which is

$\frac{d}{d x} [u v] = v \frac{d u}{d x} + u \frac{d v}{d x}$ $d/(dx) [uv] = v(du)/(dx) + u(dv)/(dx)$

where in this case

$u = {cos}^{3} x$ $u = cos^3x$
$v = {sin}^{2} x$ $v = sin^2x$ :

$= {cos}^{3} x (\frac{d}{d x} [{sin}^{2} x]) + {sin}^{2} x (\frac{d}{d x} [{cos}^{3} x))$ $= cos^3x(d/(dx)[sin^2x]) + sin^2x(d/(dx)[cos^3x))$

We can differentiate the ${sin}^{2} x$ $sin^2x$ term via the chain rule, which would look like

$\frac{d}{d x} [{sin}^{2} x] = \frac{d u^{2}}{d u} \frac{d u}{d x}$ $d/(dx)[sin^2x] = (du^2)/(du)(du)/(dx)$

where

$u = sin x$ $u = sinx$

$\frac{d}{d u} [u^{2}] = 2 u$ $d/(du) [u^2] = 2u$ (power rule):

$= {cos}^{3} x (2 \frac{d}{d x} [sin x] sin x) + {sin}^{2} x (\frac{d}{d x} [{cos}^{3} x))$ $= cos^3x(2d/(dx)[sinx]sinx) + sin^2x(d/(dx)[cos^3x))$

The derivative of $sin x$ $sinx$ is $cos x$ $cosx$ :

$= {cos}^{3} x (2 cos x sin x) + {sin}^{2} x (\frac{d}{d x} [{cos}^{3} x))$ $= cos^3x(2cosxsinx) + sin^2x(d/(dx)[cos^3x))$

Simplifying:

$= 2 {cos}^{4} x sin x + \frac{d}{d x} [{cos}^{3} x] {sin}^{2} x$ $= 2cos^4xsinx + d/(dx)[cos^3x]sin^2x$

We now use the chain rule again for differentiating the ${cos}^{3} x$ $cos^3x$ term:

$\frac{d}{d x} [{cos}^{3} x] = \frac{d u^{3}}{d u} \frac{d u}{d x}$ $d/(dx)[cos^3x] = (du^3)/(du)(du)/(dx)$

where

$u = cos x$ $u = cosx$
$\frac{d}{d u} [u^{3}] = 3 u^{2}$ $d/(du)[u^3] = 3u^2$ (power rule):

$= 2 {cos}^{4} x sin x + 3 {cos}^{2} x \frac{d}{d x} [cos x] {sin}^{2} x$ $= 2cos^4xsinx + 3cos^2xd/(dx)[cosx]sin^2x$

The derivative of $cos x$ $cosx$ is $- sin x$ $-sinx$ :

$= 2 {cos}^{4} x sin x - 3 {cos}^{2} x sin x {sin}^{2} x$ $= 2cos^4xsinx - 3cos^2xsinxsin^2x$

Simpligying gives

$= 2 {cos}^{4} x sin x - 3 {cos}^{2} x {sin}^{3} x$ $= color(blue)(2cos^4xsinx - 3cos^2xsin^3x$

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How do you find the derivative of $({sin}^{2} x) ({cos}^{3} x)$ $(sin^2x)(cos^3x)$ ?

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