How do you differentiate f(x)=csc(sqrt(2x)) f(x)=csc(2x) using the chain rule?

1 Answer
Jan 2, 2016

f'(x)=(-csc(sqrt(2x))cot(sqrt(2x)))/(sqrt(2x))

Explanation:

According to the chain rule,

d/dx(cscu)=-u'cscucotu

Thus,

f'(x)=d/dx(csc(sqrt(2x)))=-csc(sqrt(2x))cot(sqrt(2x))*d/dx(sqrt(2x))

To find d/dx(sqrt(2x)), use the chain rule again:

d/dx(sqrtu)=1/(2sqrtu)*u'

So,

d/dx(sqrt(2x))=1/(2sqrt(2x))*2=1/(sqrt(2x)

Plug this back in to find f'(x):

f'(x)=(-csc(sqrt(2x))cot(sqrt(2x)))/(sqrt(2x))