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

1 Answer
Jan 2, 2016

Answer:

#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))#