What is the derivative of #sin 5x#?

2 Answers
Apr 2, 2018

Answer:

#d/dxsin(5x)=5cos(5x)#

Explanation:

When applying the Chain Rule to trigonometric functions such as sine,

#d/dxsin(u)=cosu*(du)/dx#

In this case, #u=5x,# and so

#d/dxsin(5x)=cos(5x)*d/dx5x#

#d/dxsin(5x)=5cos(5x)#

Apr 2, 2018

Answer:

#5cos(5x)#

Explanation:

We're dealing with a composite function, and whenever we want to differentiate composite functions, we use the Chain Rule stated below:

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

Our composite function is #sin(5x)#, where:

#f(x)=sinx# and #g(x)=5x#

#f'(x)=cosx# and #g'(x)=5#

Now we just plug in! We get:

#cos(5x)*5#

Which we can rewrite as

#5cos(5x)#