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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2\[\large f(x)=\cosh^2\left(x^2\right)\]It might help to look at the function like this,\[\large f(x)=\left(\cosh x^2\right)^2\]We'll have to apply the `chain rule` several times, and this might make it easier to identity the outermost function.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2So we can see that the outermost function is the square, Taking the derivative gives us,\[\large f'(x)=2\left(\cosh x^2\right)\color{royalblue}{\left(\cosh x^2\right)'}\]We applied the power rule to the outermost function. The chain rule tells us that we have to multiply by the derivative of the inner function. The prime is to let us know that we still need to take the derivative of that part.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2\[\large f'(x)=2\left(\cosh x^2\right)\color{royalblue}{\left(\sinh x^2\right)}\color{green}{(x^2)'}\]Hopefully you recall the derivative of cosh :) So this is what we get when we take the derivative of the blue term. It leaves us with chain rule once again.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Taking the derivative of the green term gives us, \[\large f'(x)=2\left(\cosh x^2\right)\color{royalblue}{\left(\sinh x^2\right)}\color{green}{(2x)}\] Since multiplication is commutative, we can move the pieces around freely, to make it look a little nicer.\[\large f'(x)=2(2x)\left(\cosh x^2\right)\left(\sinh x^2\right)\]Which simplifies to,\[\large f'(x)=4x \cosh \left(x^2\right)\sinh \left(x^2\right)\]
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