# How do you differentiate f(x) = sin(sqrt(arccosx^2))  using the chain rule?

Jun 21, 2016

$- \frac{x \cos \left(\sqrt{\arccos {x}^{2}}\right)}{\sqrt{1 - {x}^{4}} \cdot \sqrt{\arccos {x}^{2}}}$

#### Explanation:

To differentiate $f \left(x\right)$ we have to decompose it into functions then differentiate it using chain rule:

Let:
$u \left(x\right) = \arccos {x}^{2}$
$g \left(x\right) = \sqrt{x}$
Then,
$f \left(x\right) = \sin \left(x\right)$

The derivative of the composite function using chain rule is stated as follows:

$\textcolor{b l u e}{\left(f \left(g \left(u \left(x\right)\right)\right)\right) ' = f ' \left(g \left(u \left(x\right)\right)\right) \cdot g ' \left(u \left(x\right)\right) \cdot u ' \left(x\right)}$

Let's find the derivative of each function above:

$u ' \left(x\right) = - \frac{1}{\sqrt{1 - {\left({x}^{2}\right)}^{2}}} \cdot 2 x$
color(blue)(u'(x)=-1/(sqrt(1-x^4))*2x

$g ' \left(x\right) = \frac{1}{2 \sqrt{x}}$
Subtituting $x$ by $u \left(x\right)$ we have:
color(blue)(g'(u(x))=1/(2sqrt(arccosx^2))

$f ' \left(x\right) = \cos \left(x\right)$
Substituting $x$ by $g \left(u \left(x\right)\right)$ we have to find $\textcolor{red}{g \left(u \left(x\right)\right)}$:

$\textcolor{red}{g \left(u \left(x\right)\right) = \sqrt{\arccos {x}^{2}}}$
So,
f'(g(u(x)))=cos(g(u(x))
color(blue)(f'(g(u(x)))=cos(sqrt(arccosx^2))

Substituting the calculated derivatives on the above chain rule we have:

color(blue)((f(g(u(x))))'=f'(g(u(x)))*g'(u(x))*u'(x)

$= \frac{- 2 x \cos \left(\sqrt{\arccos {x}^{2}}\right)}{2 \sqrt{1 - {x}^{4}} \cdot \sqrt{\arccos {x}^{2}}}$

$\textcolor{b l u e}{= - \frac{x \cos \left(\sqrt{\arccos {x}^{2}}\right)}{\sqrt{1 - {x}^{4}} \cdot \sqrt{\arccos {x}^{2}}}}$