How do you differentiate #y=sin(4x)#?

1 Answer
mason m
Nov 23, 2016

This is in the basic form of #sin(x)#, with the inner function changed. The derivative of #sin(x)# is #cos(x)#.

According to the chain rule, when we have a function inside another function, its derivative is the derivative of the outer function with the inside function still inside, all multiplied by the derivative of the inner function.

So, when we have some other function inside of the sine function, such as #sin(u)#, we see that its derivative will be cosine with the inner function multiplied by the derivative of the inner function—that is, #cos(u)xx# (the derivative of #u#).

Mathematically, this becomes:

#d/dx[sin(u)]=cos(u)*d/dx[u]#

So for #y=sin(4x)#, we see that:

#dy/dx=cos(4x)*d/dx[4x]=cos(4x)*4=4cos(4x)#

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