Question

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

Answer 1

`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`