How do you find the derivative of $y = 20 {sin}^{4} x$ $y=20sin^4x$ ?

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Answer 1 · 2016-08-08T07:12:14

$\frac{d y}{d x} = 80 {sin}^{3} x cos x$ $dy/dx= 80sin^3xcosx$

$y = 20 {sin}^{4} x$ $y= 20sin^4x$

$\frac{d y}{d x} = 20 \cdot 4 {sin}^{3} x \cdot \frac{d}{d x} (sin x)$ $dy/dx = 20*4 sin^3x * d/dx (sinx)$ (Power rule and Chain rule)

$= 80 {sin}^{3} x cos x$ $= 80sin^3x cosx$

Answer 2 · 2016-09-04T08:51:20

$80 {sin}^{3} (x) cos (x)$ $80 sin^(3)(x) cos(x)$

We have: $y = 20 {sin}^{4} (x)$ $y = 20 sin^(4)(x)$

This function can be differentiated using the "chain rule".

Let $u = sin(x) => u' = cos(x)$ and $v = 20 u^(4) => v' = 80 u^(3)$ :

$=> y' = cos(x) cdot 80 u^(3)$

$=> y' = 80 cos(x) u^(3)$

We can now replace $u$ with $sin(x)$ :

$=> y' = 80 cos(x) (sin(x))^(3)$

$=> y' = 80 sin^(3)(x) cos(x)$

How do you find the derivative of y=20sin^4xy=20sin4xy=20sin^4x?