How do you find the derivative of #y = e^cosh(2x)#?

1 Answer
Jun 7, 2016

Answer:

#frac{d}{dx}(e^{cosh (2x)})=e^{cosh (2x)}sinh (2x)2#

Explanation:

#frac{d}{dx}(e^{cosh (2x)})#
Applying the chain rule, #frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}#
#Let,cosh (2x)=u#
#=frac{d}{du}(e^u)frac{d}{dx}(cosh (2x))#
We know,
#frac{d}{du}(e^u)=e^u#
#frac{d}{dx}(cosh (2x))=sinh (2x)2#
So,
#frac{d}{dx}(cosh (2x))=sinh (2x)2#

substituted back,#u=cosh (2x)#

we get,

#e^{cosh (2x)}sinh (2x)2#