# What are the critical points of  f(x) = (x - 3sinx)^4 -xcosx?

Feb 26, 2018

Critical Points are where the derivative is equal to 0.

But we need to find the derivative. In order to do so, I'll split up $f \left(x\right)$ into two parts -PART 1 = ${\left(x - 3 \sin \left(x\right)\right)}^{4}$ and PART 2 = $x \cos \left(x\right)$ and I will combine them using the Sum Rule later.

PART 1:

Let's apply the Chain Rule.

$\frac{d}{\mathrm{dx}} {\left(x - 3 \sin \left(x\right)\right)}^{4} = 4 \cdot {\left(x - 3 \sin \left(x\right)\right)}^{3} \cdot \left(1 - 3 \cos \left(x\right)\right)$
$= 4 {\left(x - 3 \sin \left(x\right)\right)}^{3} \left(1 - 3 \cos \left(x\right)\right)$

PART 2:

Let's apply the Product Rule.

$\frac{d}{\mathrm{dx}} x \cos \left(x\right) = 1 \cdot \cos \left(x\right) + x \cdot - \sin \left(x\right)$
$= \cos \left(x\right) - x \sin \left(x\right)$

$\therefore f ' \left(x\right)$ is:

$= 4 {\left(x - 3 \sin \left(x\right)\right)}^{3} \left(1 - 3 \cos \left(x\right)\right) - \left(\cos \left(x\right) - x \sin \left(x\right)\right)$
$= x \sin \left(x\right) - \cos \left(x\right) + 4 {\left(x - 3 \sin \left(x\right)\right)}^{3} \left(1 - 3 \cos \left(x\right)\right)$ (Rearranging and expanding $- \left(\cos \left(x\right) - x \sin \left(x\right)\right)$)

Now all you have to do is set that equal to 0:

$x \sin \left(x\right) - \cos \left(x\right) + 4 {\left(x - 3 \sin \left(x\right)\right)}^{3} \left(1 - 3 \cos \left(x\right)\right) = 0$

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