How do you find the derivative of # h(x) = ln(cosh(2x)) #?

Question

14182 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-23T13:39:37

#h'(x)=2tanh2x#

Using the chain rule

#(dy)/dx=(dy)/(du)xx(du)/dx#

let #y=h(x)=ln(cosh(2x))#

& #u=cosh2x=>(du)/dx=2sinh2u#

so # y=lnu=>(dy)/(du)=1/u#

substitute in the chai rule

#(dy)/dx=(dy)/(du)xx(du)/dx#

#(dy)/dx=1/uxx2sinh2x#

#:. h'(x)=(2sinh2x)/(cosh2x)=2tanh2x#