derivative of trigonometry for cos to the power of 3

Question

This appears in my homework and I don't know how to do it, could you help me please?

$f(x) = \cos ^3 (4x + 1)$ with $0 < x < 1 $

Find the derivative of $f(x)$.

I know the derivative of $\cos (x)$ is equal to $-\sin (x)$, but how about $\cos(x)^3$?

Answer 1

As MJD suggests in the comments, we need to use the chain rule which basically states that: $$\text{outside}\to f\big(\underbrace{g(x)}_{\text{inside}}\big)=\left[\begin{matrix}\text{derivative }\\\text{of the outside}\\\text{evaluated at the inside}\end{matrix}\right]\times \left[\begin{matrix}\text{derivative}\\\text{of the inside}\end{matrix}\right]$$ We know that: $$\dfrac{\mathrm d}{\mathrm dx}\cos(x)=-\sin(x)\quad\color{grey}{\text{and}}\quad\dfrac{\mathrm d}{\mathrm dx}x^3=3x^2.$$ So by the chain rule: $$ \begin{align*} \dfrac{\mathrm d}{\mathrm dx}\cos^3(x)&=\dfrac{\mathrm d}{\mathrm dx}\Big[\underbrace{\cos(x)}_{\text{inside}}\Big]^{3\leftarrow\text{outside}}\\ &\overset{\text C}{\underset{\text R}=}3\big[\cos(x)\big]^2\times -\sin(x)\\\,\\ &=-3\cos^2(x)\sin(x). \end{align*} $$

In a similar fashion, you can find the derivative of basically, anything you want! Including the function you were given, and I'm sure you'll do it!

Note: With the function you were given, you have to apply the chain rule $2$ times, since you have a composition of three functions, and the formula is given as follows: $$\dfrac{\mathrm d}{\mathrm dx} f \big[g ( h(x))\big] = f'\big[g(h(x))\big]\cdot g'(h(x))\cdot h'(x). $$

Answer 2

$$f(x) = \cos^3(4x + 1) = [\cos(4x+1)]^3$$

We have a composition of three functions: $g(x) = x^3$, $\quad h(x) = \cos(4x+1)$, and $\quad q(x) = 4x+1$.

Now we can apply the chain rule: $$f'(x) = \underbrace{3[\cos(4x+1)]^2}_{g'(h(q(x)))}\cdot \underbrace{[-\sin (4x + 1)]}_{h'(q(x))}\cdot \underbrace{4}_{q'(x)}$$