MHB Derivative of Inverse Trig Function: y=4*arcsin(x/4)

[MarkFL]

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]

Hello harpazo,

We are given:

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

which implies:

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

Differentiating with respect to $y$, we obtain:

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

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

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

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

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