Differentiation problem: Find the derivative of e^{cos(3x)}

[Lez]

Dear All

The book I'm working from has a question which asks for the derivative of e^cos(3x)

I figure this is a function of a function of a function and get the answer of -3sin(3x).e^cos(3x)

The answer in the book is -sin(3x).e^cos(3x)

Can't easily see how i can lose my factor of 3.

Could someone quickly verify the correct answer?

Much appreciated
Lez

 

[ksdhart2]

Your answer is correct, and the one in the back of the book is wrong. It happens from time to time. I used a variant of your approach, involving two substitutions, but it results in the same conclusion.

Let \(\displaystyle u(x) = 3x\), so that \(\displaystyle \dfrac{du}{dx}=3\); and let \(\displaystyle v(u(x)) = cos(u(x))\), so that \(\displaystyle \dfrac{dv}{du} = -sin(u(x))\)

The problem at hand is then: \(\displaystyle \dfrac{d}{dx} \left( e^{v(u(x))} \right) = \dfrac{du}{dx} \cdot \dfrac{dv}{du} \cdot \dfrac{d}{dv} (e^v)\)

\(\displaystyle = 3 \cdot -sin(3x) \cdot e^{cos(3x)}\)

 

[Lez]

Many thanks.
Much appreciated.
Lez