# What are the derivatives of the inverse trigonometric functions?

Mar 31, 2018

See below.

#### Explanation:

$\frac{d}{\mathrm{dx}} {\sin}^{-} 1 x = \frac{1}{\sqrt{1 - {x}^{2}}}$

$\frac{d}{\mathrm{dx}} {\cos}^{-} 1 x = - \frac{1}{\sqrt{1 - {x}^{2}}}$

${\tan}^{-} 1 x = \frac{1}{1 + {x}^{2}}$

${\cot}^{-} 1 x = - \frac{1}{\sqrt{1 + {x}^{2}}}$

${\sec}^{-} 1 x = \frac{1}{x \sqrt{{x}^{2} - 1}}$

${\csc}^{-} 1 x = - \frac{1}{x \sqrt{{x}^{2} - 1}}$

One useful thing to notice is that the derivatives of all inverse "co" functions are equivalent to the derivatives of the original inverse function, but have a negative added.

Here's a proof for the derivative of the inverse sine function, if you want to avoid memorization. All the other ones can be proved in the same way.

$y = {\sin}^{-} 1 x \Leftrightarrow x = \sin y$, from the definition of an inverse function.

Differentiating $x = \sin y :$

$\frac{d}{\mathrm{dx}} \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sin y\right)$

$1 = \cos y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$ (Implicit Differentiation)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$

We need to get rid of $\cos y .$ Recall that we said $x = \sin y$ and recall the identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$. This can be rewritten for $y$ and solved for cosine as follows:

${\cos}^{2} y = 1 - {\sin}^{2} y$

$\cos y = \sqrt{1 - {\sin}^{2} y}$

Recalling that $x = \sin y ,$ then ${\sin}^{2} y = {x}^{2}$

Thus,

$\cos y = \sqrt{1 - {x}^{2}}$

$\frac{d}{\mathrm{dx}} {\sin}^{-} 1 x = \frac{1}{\sqrt{1 - {x}^{2}}}$

I recommend that you commit these integrals to memory- they will help you when you are learning the trig. substitution.