How do you differentiate #f(x)=e^(csc2x)# using the chain rule.?

Question

1356 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-14T11:37:52

#f'(x) = -2e^csc(2x)csc(2x)cot(2x)#

Using the chain rule, we get

#f'(x) = d/dxe^(csc(2x))#

#= e^(csc(2x))(d/dxcsc(2x))#

#= e^(csc(2x))(-csc(2x)cot(2x))(d/dx2x)#

#=e^(csc(2x))(-csc(2x)cot(2x))(2)#

Thus

#f'(x) = -2e^csc(2x)csc(2x)cot(2x)#