# How do you differentiate y=arcsin(x)?

## I don't even know where to start. I watched a video on You Tube of how to do it, but it just didn't make any sense to me. Hope someone can help. Thanks

Jan 17, 2018

See below.

#### Explanation:

$y = \arcsin \left(x\right)$

Before we proceed we need to understand just what it is we are looking for. Remember that:

$y = \arcsin \left(x\right)$ is the inverse function of $y = \sin \left(x\right)$

This can be expressed as:

$y = \arcsin \left(x\right) \iff x = \sin \left(y\right)$

Using

$x = \sin \left(y\right)$

We need to differentiate in respect of $x$, so this will need to be differentiated implicitly.

Remembering that:

$\frac{d}{\mathrm{dx}} \left(f \left(y\right)\right) = \frac{d}{\mathrm{dy}} \left(f \left(y\right)\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(x\right) = \frac{d}{\mathrm{dy}} \left(\sin \left(y\right)\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$1 = \cos \left(y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(y\right)$

From above $y = \arcsin \left(x\right)$

Substitutng in $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(y\right)$

$\therefore$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(\arcsin \left(x\right)\right)$

This is a little awkward, and it would be easier if we could express this in a different way.

Using the Pythagorean identity:

$\textcolor{red}{{\sin}^{2} \left(y\right) + {\cos}^{2} \left(y\right) = 1}$

$\cos \left(y\right) = \pm \sqrt{1 - {\sin}^{2} \left(y\right)}$

Using positive root: ( see below)

$\therefore$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\sin}^{2} \left(y\right)}}$

From above $x = \sin \left(y\right)$

Hence:

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

$\therefore$

dy/dx(arcsin(x))=1/(sqrt(1-x^2)

Reason for using positive root of $\sqrt{1 - {\sin}^{2} \left(y\right)}$

This is because $y = \sin \left(x\right)$ only has an inverse if we restrict the domain to:

$- \frac{\pi}{2} \le x \le \frac{\pi}{2}$

In $\textcolor{w h i t e}{88} \frac{1}{\cos} \left(y\right)$ , $\textcolor{w h i t e}{88} y$ is an angle and it is an angle in the range

$- \frac{\pi}{2} \le x \le \frac{\pi}{2}$

This range is in the I and IV quadrants, where the cosine ratios are positive.

Jan 17, 2018

Here's an alternative solution

#### Explanation:

This troubled me as well when I was first learning these kinds of derivatives. Here's how I learned how to tackle this type of problem:

Similarly to what was said below, $y = \arcsin \left(x\right)$ is just the inverse of $y = \sin \left(x\right)$

When trying to take the derivative of $y = \arcsin \left(x\right)$ we can rewrite this as:

color(blue)(sin(y)=x

It is at this point where I would create a right triangle showing this

Recall: $\sin \left(y\right) = \text{Opposite"/"Hypothenuse}$

In this case $\sin \left(y\right) = \frac{x}{1}$

We can solve for the missing side using ${a}^{2} + {b}^{2} = {c}^{2}$ which will give you $\sqrt{1 - {x}^{2}}$. This will be useful later!

Now we'll differentiate implicitly (W.R.T $x$):

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \cos \left(y\right) = 1$

Divide $\cos \left(y\right)$ from both sides

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(y\right)$

Now we'll have to rewrite in terms of $x$ but you may notice that we $\textcolor{red}{\cos \left(y\right)}$ in the way but what is color(red)(cos(y)?

Well based on the triangle we made above:

$\cos \left(y\right) = \text{Adjacent"/"Hypothenuse}$

color(red)(cos(y)=sqrt(1-x^2)/1=sqrt(1-x^2)

So $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(y\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left[\arcsin \left(x\right)\right] = \frac{1}{\sqrt{1 - {x}^{2}}}$

The neat thing about this is that this is useful for the other inverse functions like $y = \arccos \left(x\right)$

$y = \arccos \left(x\right) \iff \cos \left(y\right) = x$

Create a triangle:

Differentiate both sides implicitly (W.R.T $x$)

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot - \sin \left(y\right) = 1$

Divide both sides by $- \sin \left(y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} \left(y\right)$

Rewrite in terms of $x$

Since $\cos \left(y\right) = \frac{x}{1}$

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

So $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} \left(y\right) = - \frac{1}{\sqrt{1 - {x}^{2}}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left[\arccos \left(x\right)\right] = - \frac{1}{\sqrt{1 - {x}^{2}}}$

Also if you have something like $y = \arcsin \left(3 x\right)$ the same process applies but the triangle will be different based on whatever the coefficient is.

So...

$y = \arcsin \left(3 x\right) \iff \sin \left(y\right) = 3 x$

Differentiate implicitly (W.R.T $x$):

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \cos \left(y\right) = 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\cos} \left(y\right)$

Since $\sin \left(y\right) = \frac{3 x}{1}$

Then $\cos \left(y\right) = \frac{\sqrt{1 - 9 {x}^{2}}}{1} = \sqrt{1 - 9 {x}^{2}}$

So $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\cos} \left(y\right) = \frac{3}{\sqrt{1 - 9 {x}^{2}}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left[\arcsin \left(3 x\right)\right] = \frac{3}{\sqrt{1 - 9 {x}^{2}}}$

I hope I explained this well!

Jan 17, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(\arcsin x\right) = \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

#### Explanation:

There is another way to do it. We have a formula for calculating the derivatives of inverse functions:

(f^-1)^'(x)=1/(f^'(f^-1(x))

Here ${f}^{-} 1 \left(x\right) = \arcsin x \Rightarrow f \left(x\right) = \sin x$ and ${f}^{'} \left(x\right) = \cos x$

So ${\left(\arcsin x\right)}^{'} = \frac{1}{\cos} \left(\arcsin x\right)$

Using the arguments given in the other answers, we know that

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