Question

What is the equation of the tangent line of `f(x) = sin^-1x/x` at `x=pi/4`?

Answer 1

To get the tangent of an equation at a point, we need to differentiate the equation and then use the derivative as a slope, substituting that value into the general slope equation.

So first, (assuming you do know how to differentiate according to quotient rule)
`\frac{d}{dx}(f(x))=\frac{x/\sqrt{1-x^2}-sin^{-1}(x)}{x^2}`
Substituting for `x=sin(\theta)\implies\sin^{-1}(x)=\theta` we get `x/\sqrt{1-x^2}=tan(sin^{-1}(x))`
So `\frac{d}{dx}(f(x))=\frac{tan(sin^{-1}(x))-sin^{-1}(x)}{x^2}`
At, `x=\pi/4 " The slope is " \frac{tan(sin^{-1}(\pi/4))-sin^{-1}(\pi/4)}{(\pi/4)^2}`

Now, before we proceed, we need to simplify the slope equation first.
The given slope equation is `y-y_o=m(x-x_o)`
From the above case, we know the tangent passes through the point `(\pi/4,4/(\pi\sqrt{2}))` So `x_o=\pi/4` and `y_o=4/(\pi\sqrt{2})`
So the equation for the tangent that occurs at `x=\pi/4` for the function `f(x)=sin^{-1}(x)/x` is
`y-4/(\pi\sqrt{2})=\frac{tan(sin^{-1}(\pi/4))-sin^{-1}(\pi/4)}{(\pi/4)^2}(x-\pi/4)`

That's as simple as I can get it.