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

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Answer 1 · 2016-08-08T07:12:14

#dy/dx= 80sin^3xcosx#

#y= 20sin^4x#

#dy/dx = 20*4 sin^3x * d/dx (sinx) # (Power rule and Chain rule)

#= 80sin^3x cosx#

Answer 2 · 2016-09-04T08:51:20

#80 sin^(3)(x) cos(x)#

We have: #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)#