# How do you find the derivative of sin(arccosx)?

Aug 1, 2015

Use the Chain Rule, resulting in $\cos \left({\cos}^{-} 1 \left(x\right)\right) \cdot \frac{- 1}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

The Chain Rule involves breaking a function up into a function of a function, sometimes expressed as $y = f \left(u\right)$ and $u = g \left(x\right)$.

Step 1. Let $y = \sin \left(u\right)$ and $u = {\cos}^{-} 1 \left(x\right)$.

Step 2. Chain Rule says $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Step 3. Apply the Chain Rule to the equations in Step 1.

$\frac{\mathrm{dy}}{\mathrm{du}} = \cos \left(u\right)$ and (du)/(dx)=(-1)/(sqrt(1-x^2).

Step 4. Get rid of the $u$'s in $\frac{\mathrm{dy}}{\mathrm{du}}$. Recall that $u = {\cos}^{-} 1 \left(x\right)$, so then

$\frac{\mathrm{dy}}{\mathrm{du}} = \cos \left({\cos}^{-} 1 \left(x\right)\right)$

Step 5. Plug your derivatives back into Chain Rule of Step 2.

dy/dx=(dy)/(du)*(du)/(dx)=cos(cos^-1(x))*(-1)/(sqrt(1-x^2)

Step 6 is often optional, but you can then try to simplify as much as you can.

dy/dx=(dy)/(du)*(du)/(dx)=(-cos(cos^-1(x)))/(sqrt(1-x^2).

Aug 1, 2015

I would use the fact that $\sin \left(\arccos x\right) = \sqrt{1 - {x}^{2}}$

#### Explanation:

$\arccos x$ is an angle (or, a number) between $0$ and $\pi$ whose cosine is $x$.
The sine of an angle (or, a number) between $0$ and $\pi$ whose cosine is $x$ is $\sqrt{1 - {x}^{2}}$.
(Use ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$. So $\cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$
and with $\theta$ in $\left[0 , \pi\right]$, we have $\cos \theta > 0$.)

$\sin \left(\arccos x\right) = \sqrt{1 - {x}^{2}}$

Now i'll use $\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \frac{\mathrm{du}}{\mathrm{dx}}$ to get:

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

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

$= \frac{- x}{\sqrt{1 - {x}^{2}}}$