What is the derivative of #sin(4x)#?

1 Answer
Jun 25, 2018

#4cos(4x)#

Explanation:

We are dealing with a composite function here, so it helps to use the Chain Rule

#f'(g(x))*g'(x)#

Our function is essentially in the form #f(g(x))#

where #f(x)=sinx# and #g(x)=4x#

Let's find the derivatives and plug into the Chain Rule:

#f'(x)=cosx#, #g'(x)=4#

We have both of our functions and derivatives, now let's plug in:

#cos(4x)*4#

#color(steelblue)(4cos(4x))#

Hope this helps!