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

Jan 25, 2018

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

#### Explanation:

Given: $f \left(x\right) = \sin \left(\sqrt{\arcsin} x\right)$

We'll use the chain rule but to make this easier to visualize we are going to let $\sqrt{\arcsin x} = u$

So we have

$f \left(x\right) = \sin \left(u\right)$

By the chain rule, we are solving

f'(x)=d/dx[sin(sqrtarcsinx)]=color(red)((df)/(du)[sin(u)])*color(blue)((du)/dx[sqrtarcsinx]

The derivative of color(red)(sin(u) is $\textcolor{red}{\cos \left(u\right)}$

:.color(red)((df)/dx[sin(u)]=cos(u)

For color(blue)((du)/dx[sqrtarcsinx] we can use the chain rule again.

We can call $\sqrt{\arcsin} x$ a new function, let's call it $g \left(x\right)$

g(x)=sqrtarcsinx=>color(blue)(g'(x)=(dg)/dx[sqrtarcsinx]=(du)/dx[sqrtarcsinx]

And similarly we'll let $\arcsin x = v$

So we have

$g \left(x\right) = \sqrt{v}$

By the chain rule we are differentiating

color(blue)((dg)/dx[sqrtarcsinx])=color(purple)((dg)/(dv)[sqrtv]*color(green)((dv)/dx[arcsinx]

For color(purple)((dg)/(dv)[sqrtv]=>d/dx[v^(1/2)]

By the Power Rule

color(purple)((dg)/(dv)=1/2v^(1/2-1)=1/2v^(-1/2)=1/(2v^(1/2))=1/(2sqrtv)

:.color(purple)((dg)/(dv)[sqrtv]=1/(2sqrtv)

For color(green)((dv)/dx[arcsinx], it's derivative is color(green)(1/sqrt(1-x^2)

:.color(green)((dv)/dx[arcsinx]=1/sqrt(1-x^2)

Now that we know what color(purple)((dg)/(dv) and color(green)((dv)/dx are, we can find $\textcolor{b l u e}{\frac{\mathrm{dg}}{\mathrm{dx}} \left[\sqrt{\arcsin} x\right]}$

$\textcolor{b l u e}{\frac{\mathrm{dg}}{\mathrm{dx}} \left[\sqrt{\arcsin} x\right]} = \textcolor{p u r p \le}{\frac{\mathrm{dg}}{\mathrm{dv}} \left[\sqrt{v}\right]} \cdot \textcolor{g r e e n}{\frac{\mathrm{dv}}{\mathrm{dx}} \left[\arcsin x\right]} = \frac{1}{2 \sqrt{v}} \cdot \frac{1}{\sqrt{1 - {x}^{2}}} = \frac{1}{2 \sqrt{v} \sqrt{1 - {x}^{2}}}$

Okay, now let us summarize

$u = \sqrt{\arcsin} x \text{ } v = \arcsin x$

color(red)((df)/dx[sin(u)]=cos(u) " " color(blue)((du)/dx[sqrtarcsinx]=1/(2sqrtvsqrt(1-x^2)))

Going back to the original problem:

$f ' \left(x\right) = \textcolor{red}{\frac{\mathrm{df}}{\mathrm{du}} \left[\sin \left(u\right)\right]} \cdot \textcolor{b l u e}{\frac{\mathrm{du}}{\mathrm{dx}} \left[\sqrt{\arcsin} x\right]} = \textcolor{red}{\cos \left(u\right)} \cdot \textcolor{b l u e}{\frac{1}{2 \sqrt{v} \sqrt{1 - {x}^{2}}}}$

Now we have to substitute in back the values for $u$ and $v$ which will give us the final result:

$f ' \left(x\right) = \textcolor{red}{\cos \left(\sqrt{\arcsin} x\right)} \cdot \textcolor{b l u e}{\frac{1}{2 \sqrt{\arcsin} x \sqrt{1 - {x}^{2}}}}$

$f ' \left(x\right) = \frac{\textcolor{red}{\cos \left(\sqrt{\arcsin} x\right)}}{\textcolor{b l u e}{\left(2 \sqrt{\arcsin} x \sqrt{1 - {x}^{2}}\right)}}$

I hope this helped! :)