# How do I find the derivative of y = csc^-1(x^2+1)?

Jun 8, 2015

There is a pattern amongst the inverse trig derivatives:

$\frac{d}{\mathrm{dx}} \left[\arcsin u\right] = \frac{1}{\sqrt{1 - {u}^{2}}} \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

$\frac{d}{\mathrm{dx}} \left[\arccos u\right] = - \frac{1}{\sqrt{1 - {u}^{2}}} \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

$\frac{d}{\mathrm{dx}} \left[a r c \sec u\right] = \frac{1}{| u | \sqrt{{u}^{2} - 1}} \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

$\frac{d}{\mathrm{dx}} \left[a r c \csc u\right] = - \frac{1}{| u | \sqrt{{u}^{2} - 1}} \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

$\frac{d}{\mathrm{dx}} \left[\arctan u\right] = \frac{1}{1 + {u}^{2}} \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

$\frac{d}{\mathrm{dx}} \left[a r c \cot u\right] = - \frac{1}{1 + {u}^{2}} \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

Notice though that Wolfram Alpha would say that the derivative of $a r c \csc u$ is -1/(sqrt(1-1/(u^2))*u^2, but if you use its answer and manipulate it by calling the outer ${u}^{2}$ equal to $| u | | u | = | u | \cdot \sqrt{{u}^{2}}$ and distributing it into the square root, it will be equivalent. We can use both, though.

$\frac{d}{\mathrm{dx}} \left[a r c \csc \left({x}^{2} + 1\right)\right] = - \frac{1}{| {x}^{2} + 1 | \sqrt{{\left({x}^{2} + 1\right)}^{2} - 1}} \cdot \left(2 x\right)$

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

or...

$\frac{d}{\mathrm{dx}} \left[a r c \csc \left({x}^{2} + 1\right)\right] = - \frac{1}{{\left({x}^{2} + 1\right)}^{2} \sqrt{1 - \frac{1}{{x}^{2} + 1} ^ 2}} \cdot \left(2 x\right)$

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