Trig Derivative Triple

[Jason76]

\(\displaystyle cos(cos(cos(x)))\)

Wouldn't this be \(\displaystyle sin(x)\)

\(\displaystyle \dfrac{d}{dx} cos(x) = -sin(x)\)

\(\displaystyle \dfrac{d}{dx} -sin(x) = -cos(x)\)

\(\displaystyle \dfrac{d}{dx} -cos(x) = sin(x)\) Online homework says this is wrong though.

 

[stapel]

\(\displaystyle cos(cos(cos(x)))\)

Wouldn't this be \(\displaystyle sin(x)\)

\(\displaystyle \dfrac{d}{dx} cos(x) = -sin(x)\)

\(\displaystyle \dfrac{d}{dx} -sin(x) = -cos(x)\)

\(\displaystyle \dfrac{d}{dx} -cos(x) = sin(x)\) Online homework says this is wrong though.

How do your three derivatives (of single trig functions) relate to the original compositional function? At what stage did you apply the Chain Rule, and how?

 

[HallsofIvy]

What you did was take the third derivative of cos(x). What you are asked to do is take the first derivative of y= cos(cos(cos(x))).

Let u= cos(cos(x)). Then y= cos(u). Let v= cos(x). Then u= cos(v).

\(\displaystyle \frac{dy}{dx}=\)\(\displaystyle \left(\frac{dy}{du}\right)\)\(\displaystyle \left(\frac{du}{dx}\right)=\)\(\displaystyle \left(\frac{dy}{du}\right)\)\(\displaystyle \left(\frac{du}{dv}\right)\)\(\displaystyle \left(\frac{dv}{dx}\right)\)