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How do you use the chain rule to differentiate #y=sin^5x#?

1 Answer

Sep 25, 2016

The chain rule is #d[f(u(x))]/(dx) = (df)/(du)(du)/(dx)# so let #u(x) = sin(x)# then #f(u) = u^5# and it becomes quite simple.

Explanation:

#(df)/(du) = 5u^4#

#(du)/(dx) = cos(x)#

#(df)/(du)(du)/(dx) = 5u^4cos(x)#

Reverse the u substitution:

#(df)/(du)(du)/(dx) = 5sin^4(x)cos(x)#

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