Question #91477

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Question #91477

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Answer 1 · 2018-01-07T13:16:42

$3 {sin}^{2} (x) cos (x)$ $3sin^2(x)cos(x)$

Explanation:

${sin}^{3} (x) \Leftrightarrow {(sin x)}^{3}$ $sin^3(x)<=>(sinx)^3$

Using the chain rule:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du)*(du)/dx$

Let $u = sin (x)$ $u=sin(x)$

$\frac{d y}{d x} ({sin}^{3} (x)) = \frac{d y}{d u} {(u)}^{3} \cdot \frac{d u}{d x} (u)$ $dy/dx(sin^3(x))=dy/(du)(u)^3*(du)/dx(u)$

$88888888888 = 3 u^{2} \cdot cos (x)$ $color(white)(88888888888)=3u^2*cos(x)$

$u = sin (x)$ $u=sin(x)$

$88888888888 = 3 {sin}^{2} (x) cos (x)$ $color(white)(88888888888)=3sin^2(x)cos(x)$

Answer 2 · 2018-01-07T13:18:10

$3 {sin}^{2} (x) cos (x)$ $3sin^2(x)cos(x)$

Explanation:

Using the chain rule,

$\frac{d f (u)}{d x} = \frac{d f}{d u} \cdot \frac{d u}{d x}$ $(df(u))/dx=(df)/(du)*(du)/(dx)$

Let $u = sin (x)$ $u=sin(x)$

$\frac{d u}{d x} = cos (x)$ $(du)/(dx)=cos(x)$

$f = u^{3}$ $f=u^3$

$\frac{d f}{d u} = 3 u^{2}$ $(df)/(du)=3u^2$

$:.(df(u))/dx=3u^2cos(x)$

Substitute $u=sin(x)$ back, we get

$:.(df(u))/dx=3sin^2(x)cos(x)$

Answer 3 · 2018-01-07T13:25:41

$3sin^2xcosx$

Explanation:

$"differentiate using the "color(blue)"chain rule"$

$"given "y=f(g(x))" then"$

$dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"$

$"express "sin^3x=(sinx)^3$

$rArrd/dx((sinx)^3)$

$=3(sinx)^2xxd/dx(sinx)=3sin^2xcosx$

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Question #91477

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