# Differentiating Inverse Trigonometric Functions

Derivatives of the Inverse Trig Functions

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• Most people remember this
$f ' \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$
as one of derivative formulas; however, you can derive it by implicit differentiation.

Let us derive the derivative.
Let $y = {\sin}^{- 1} x$.

By rewriting in terms of sine,
$\sin y = x$

By implicitly differentiating with respect to $x$,
$\cos y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

By dividing by $\cos y$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$

By $\cos y = \sqrt{1 - {\sin}^{2} y}$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\sin}^{2} y}}$

By $\sin y = x$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$

• By Implicit Differentiation,
$y ' = - \frac{1}{\sqrt{1 - {x}^{2}}}$.

Let us look at some details.
$y = {\cos}^{- 1} x$

by rewriting in term of cosine,
$R i g h t a r r o w \cos y = x$

by implicitly differentiating with respect to $x$,
$R i g h t a r r o w - \sin y \cdot y ' = 1$

by dividing by $- \sin y$,
$y ' = - \frac{1}{\sin} y$

by the trionometric identity $\sin y = \sqrt{1 - {\cos}^{2} y}$,
$y ' = - \frac{1}{\sqrt{1 - {\cos}^{2} y}}$

by $\cos y = x$,
$y ' = - \frac{1}{\sqrt{1 - {x}^{2}}}$

• I seem to recall my professor forgetting how to deriving this. This is what I showed him:

$y = \arctan x$

$\tan y = x$

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} y}$

Since $\tan y = \frac{x}{1}$ and $\sqrt{{1}^{2} + {x}^{2}} = \sqrt{1 + {x}^{2}}$, ${\sec}^{2} y = {\left(\frac{\sqrt{1 + {x}^{2}}}{1}\right)}^{2} = 1 + {x}^{2}$

$\implies \textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}}$

I think he originally intended to do this:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} y}$

${\sec}^{2} y = 1 + {\tan}^{2} y$

${\tan}^{2} y = x \to {\sec}^{2} y = 1 + {x}^{2}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$

• The answer is $y ' = - \frac{1}{1 + {x}^{2}}$

We start by using implicit differentiation:

$y = {\cot}^{- 1} x$
$\cot y = x$
$- {\csc}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{{\csc}^{2} y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {\cot}^{2} y}$ using trig identity: $1 + {\cot}^{2} \theta = {\csc}^{2} \theta$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {x}^{2}}$ using line 2: $\cot y = x$

The trick for this derivative is to use an identity that allows you to substitute $x$ back in for $y$ because you don't want leave the derivative as an implicit function; substituting $x$ back in will make the derivative an explicit function.

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