# What is the derivative of this function y=csc^-1(4x^2)?

Dec 27, 2017

dy/dx=-4/(xsqrt(16x^4-1)

#### Explanation:

We first need to work the derivative of ${\csc}^{-} 1 \left(x\right)$

$y = {\csc}^{-} 1 \left(x\right) \to \csc \left(y\right) = x$

Now differentiate both sides implicitly with respect to $x$ to get:

$- \frac{\mathrm{dy}}{\mathrm{dx}} \csc \left(y\right) \cot \left(y\right) = 1$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\csc \left(y\right) \cot \left(y\right)}$

Now, using:

${\sin}^{2} \left(y\right) + {\cos}^{2} \left(y\right) = 1$

Divide this identity through by ${\sin}^{2} \left(y\right)$ to get:

$1 + {\cot}^{2} \left(y\right) = {\csc}^{2} \left(y\right) \to \cot \left(y\right) = \sqrt{{\csc}^{2} \left(y\right) - 1}$

We can now plug this into our equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$ to get:

dy/dx = -1/(csc(y)sqrt(csc^2(y)-1)

Using $\csc \left(y\right) = x$ we may now rewrite $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $x$:

dy/dx = -1/(xsqrt(x^2-1)

So now that we have the derivative of ${\csc}^{- 1} x$ we can now apply the chain rule to obtain the derivative of $y = {\csc}^{-} 1 \left(4 {x}^{2}\right)$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\left(4 {x}^{2}\right) \sqrt{{\left(4 {x}^{2}\right)}^{2} - 1}} . \frac{d}{\mathrm{dx}} \left(4 {x}^{2}\right)$

$= - \frac{8 x}{\left(4 {x}^{2}\right) \sqrt{{\left(4 {x}^{2}\right)}^{2} - 1}}$

Now simplify:

=-4/(xsqrt(16x^4-1)