# How do you find the derivative of y = Arcsin ((3x)/4)?

Feb 11, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\sqrt{16 - \left(9 {x}^{2}\right)}}$

#### Explanation:

You'll need to use the chain rule. Recall that the formula for this is:

$f \left(g \left(x\right)\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

The idea is that you take the derivative of the outermost function first, and then just work your way inside.

Before we start, let's identify all our functions in this expression. We have:

• $\arcsin \left(x\right)$

• $\frac{3 x}{4}$

$\arcsin \left(x\right)$ is the outermost function, so we'll start by taking the derivative of that. So:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[\arcsin \left(3 \frac{x}{4}\right)\right] = \frac{1}{\sqrt{1 - {\left(\frac{3 x}{4}\right)}^{2}}}}$

Notice how we're still preserving that $\left(\frac{3 x}{4}\right)$ in there. Remember, when using the chain rule you differentiate outside-in, but you still keep the inner functions when differentiating the outer ones.

$\frac{3 x}{4}$ is our next outermost function, so we'll need to tag the derivative of that as well. So:

$\textcolor{g r e y}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[\arcsin \left(3 \frac{x}{4}\right)\right] = \frac{1}{\sqrt{1 - {\left(\frac{3 x}{4}\right)}^{2}}}} \cdot \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(\frac{3 x}{4}\right)}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\left(\frac{3 x}{4}\right)}^{2}}} \cdot \left(\frac{3}{4}\right)$

And that's the end of the calculus portion to this problem! All that's left is to do some simplification to tidy up this expression, and we end up with:

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\sqrt{16 - \left(9 {x}^{2}\right)}}$

If you'd like some additional help on the Chain Rule, I'd encourage you to take a look at some of my videos on the subject:

Hope that helped :)

Feb 11, 2018

Given: color(blue)(y=f(x)=sin^(-1) ((3x)/4)

color(green)(d/(dx)[sin^(-1) ((3x)/4)]=3/sqrt(16-9x^2)

#### Explanation:

Given:

color(blue)(y=f(x)= sin^(-1) ((3x)/4)

Function Composition is applying one function to the results of another:

Observe that the argument of the trigonometric function ${\sin}^{- 1} \left(\text{ }\right)$ is also a function.

The Chain Rule is a rule for differentiating compositions of functions like the one that we have.

Chain Rule:

color(red)(dy/(dx) = (dy/(du))*((du)/(dx))" " (or)

color(blue)(d/(dx)[f{g(x)}] = f'[g(x)]*g'[x]

We are given

color(blue)(y=f(x)= sin^(-1) ((3x)/4)

Let,

$f \left(x\right) = {\sin}^{- 1} \left(u\right) \text{ "and " } u = \frac{3 x}{4}$

color(green)(Step.1

We will differentiate

$f \left(x\right) = {\sin}^{- 1} \left(u\right) \text{ }$ Function.1

using the common derivative result:

color(brown)(d/(dx)sin^(-1)(x) = 1/sqrt(1-x^2

Using the above result we can differentiate Function.1 above as

$\frac{d}{\mathrm{du}} {\sin}^{- 1} \left(u\right) = \frac{1}{\sqrt{1 - {u}^{2}}} \text{ }$ Result.1

color(green)(Step.2

In this step, we will differentiate the inside function $\frac{3 x}{4}$

$\frac{d}{\mathrm{dx}} \left(\frac{3 x}{4}\right)$

Pull the constant out

$\Rightarrow \frac{3}{4} \cdot \frac{d}{\mathrm{dx}} \left(x\right)$

$\Rightarrow \frac{3}{4} \cdot 1$

$\Rightarrow \frac{3}{4}$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{3 x}{4}\right) = \frac{3}{4} \text{ }$Result.2

color(green)(Step.3

We will use the two intermediate results, Result.1 and Result.2 to proceed.

color(green)(d/(dx)[sin^(-1) ((3x)/4)] = 1/sqrt(1-u^2)*(3/4)

Substitute back color(brown)(u=((3x)/4)

Then,

color(green)(d/(dx)[sin^(-1) ((3x)/4)] = 1/sqrt(1-((3x)/4)^2)*(3/4)

$\Rightarrow \left(\frac{3}{4}\right) \cdot \frac{1}{\sqrt{1 - {\left(\frac{3 x}{4}\right)}^{2}}}$

$\Rightarrow \left(\frac{3}{4}\right) \cdot \frac{1}{\sqrt{1 - \left(\frac{9 {x}^{2}}{16}\right)}}$

$\Rightarrow \left(\frac{3}{4}\right) \cdot \frac{1}{\sqrt{\frac{16 - 9 {x}^{2}}{16}}}$

$\Rightarrow \left(\frac{3}{4}\right) \cdot \frac{1}{\sqrt{\frac{16 - 9 {x}^{2}}{{4}^{2}}}}$

rArr (3/4)*1/(sqrt((16-9x^2))/(sqrt((4^2)))

$\Rightarrow \left(\frac{3}{4}\right) \cdot \frac{1}{\sqrt{\left(16 - 9 {x}^{2}\right)}} \cdot 4$

$\Rightarrow \left(\frac{3}{\cancel{4}}\right) \cdot \frac{1}{\sqrt{\left(16 - 9 {x}^{2}\right)}} \cdot \cancel{4}$

$\Rightarrow \frac{3}{\sqrt{\left(16 - 9 {x}^{2}\right)}}$

Hence, our final answer can be written as

color(green)(d/(dx)[sin^(-1) ((3x)/4)]=3/sqrt(16-9x^2)