# How do you find the derivative of arccos(sinx^3)?

Oct 10, 2016

Use the chain rule, twice

$\frac{d \left[{\cos}^{-} 1 \left(\sin \left({x}^{3}\right)\right)\right]}{\mathrm{dx}} = - 3 {x}^{2}$

#### Explanation:

The chain rule is:

$\frac{d \left[f \left(g \left(x\right)\right)\right]}{\mathrm{dx}} = \left(\frac{\mathrm{df} \left(g\right)}{\mathrm{dg}}\right) \left(\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}\right)$

Let $g \left(x\right) = \sin \left({x}^{3}\right)$, then $f \left(g\right) = {\cos}^{-} 1 \left(g\right)$ and (df(g))/(dg) = -1/sqrt(1 - g^2

For $\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}$, use the chain rule, again:

$\frac{d \left[g \left(h \left(x\right)\right)\right]}{\mathrm{dx}} = \left(\frac{\mathrm{dg} \left(h\right)}{\mathrm{dh}}\right) \left(\frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}}\right)$

Let $h \left(x\right) = {x}^{3}$, then $g \left(h\right) = \sin \left(h\right)$, $\frac{\mathrm{dg} \left(h\right)}{\mathrm{dh}} = \cos \left(h\right)$, and $\frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}} = 3 {x}^{2}$

Substitute back into the second chain rule:

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = 3 {x}^{2} \cos \left(h\right)$

Reverse the substitution for h:

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = 3 {x}^{2} \cos \left({x}^{3}\right)$

Substitute back into the first chain rule:

$\frac{d \left[f \left(g \left(x\right)\right)\right]}{\mathrm{dx}} = - 3 {x}^{2} \cos \frac{{x}^{3}}{\sqrt{1 - {g}^{2}}}$

Reverse the substitution for g:

$\frac{d \left[{\cos}^{-} 1 \left(\sin \left({x}^{3}\right)\right)\right]}{\mathrm{dx}} = - 3 {x}^{2} \cos \frac{{x}^{3}}{\sqrt{1 - {\sin}^{2} \left({x}^{3}\right)}}$

Please observe that $\sqrt{1 - {\sin}^{2} \left({x}^{3}\right)} = \cos \left({x}^{3}\right)$ which causes the fraction to become 1:

$\frac{d \left[{\cos}^{-} 1 \left(\sin \left({x}^{3}\right)\right)\right]}{\mathrm{dx}} = - 3 {x}^{2}$