What is the derivative of #sin x^5#?

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Answer 1 · 2018-06-10T14:48:01

#dy/dx=cos(x^5)*5x^4#

#y=sinx^5#

Using the chain rule:

If #y=y(u)# and #u=u(x)# then, #dy/dx=(dy/(du))*((du)/dx)#

#y=sinu#
#dy/dx=cosu#

#u=x^5#
#du/dx=5x^4#

Therefore,
#dy/dx=(dy/(du))*((du)/dx)#
#dy/dx=cosu*5x^4#

Substitute #u# back in and rearrange:
#dy/dx=cos(x^5)*5x^4#