# harpazo's question at Yahoo! Answers regarding the derivative of an inverse trigonometric function

#### MarkFL

Staff member
Here is the question:

Derivative of Inverse Function?

Find d/dx of y = 4*arcsin(x/4)
I have posted a link there to this thread so the OP can see my work.

#### MarkFL

Staff member
Hello harpazo,

We are given:

$$\displaystyle y=4\sin^{-1}\left(\frac{x}{4} \right)$$

which implies:

$$\displaystyle \sin\left(\frac{y}{4} \right)=\frac{x}{4}$$

Differentiating with respect to $y$, we obtain:

$$\displaystyle \frac{1}{4}\cos\left(\frac{y}{4} \right)=\frac{1}{4}\frac{dx}{dy}$$

Solving for $$\displaystyle \frac{dy}{dx}$$, we get:

$$\displaystyle \frac{dy}{dx}=\sec\left(\frac{y}{4} \right)$$

Since $$\displaystyle \frac{y}{4}=\sin^{-1}\left(\frac{x}{4} \right)$$ we have:

$$\displaystyle \frac{dy}{dx}=\sec\left(\sin^{-1}\left(\frac{x}{4} \right) \right)=\frac{4}{\sqrt{16-x^2}}$$