What is the equation for a tangent to the graph of $y=\arcsin(x/2)$ at the origin?

Question

I believe arc sin is the same as inverse sin but then I don't know how to deal with taking the derivative of that.

Answer 1

Hint To take derivatives, take $\sin$ of both sides: $\sin y = x/2$ and now differentiate implicitly.

EDIT In response, let me show how implicit differentiation would work. Differentiate both sides of $$\sin y = x/2$$ with respect to $x$ to get $$y'\cos y = 1/2.$$ Now use the original relationship between $x$ and $y$ to get $$ y' = \frac{1}{2\cos y} = \frac{1}{2\sqrt{1 - \sin^2 y}} = \frac{1}{2\sqrt{1 - (x/2)^2}} = \frac{1}{2\sqrt{1 - x^2/4}} = \frac{1}{\sqrt{4 - x^2}} $$

Answer 2

Use the formula for the derivative of the inverse:

$$ \arcsin ' x = \frac 1{\cos(\arcsin x)} = \frac 1{\sqrt{1-x^2}} $$ and then take the composition with $x\to x/2$:

$$ \frac d{dx} \left(\arcsin \frac x2 \right)= \frac 12 \frac 1{\sqrt{1-x^2/4}} $$ Take $x=0$ and you get the equation of the tangent line: $$ T(x) = \arcsin 0 + \left.\frac d{dx} \left(\arcsin \frac x2 \right) \right|_{x=0} x =\frac x2 $$