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Differentiation of the Inverse Trigonometric Functions

April 8th, 2009 |

Author: Merrin

Based on the inverse theory of differentiation we can now find the derivatives of the inverse trigonometric functions. These functions are written with an “arc” preceding them to indicate they are inverses. At several spots in the derivation we will be forced to choose a plus or minus sign. This is best determined by determining the sign of the slope of the functions using a graph. Graphing the inverse functions only requires knowledge of the functions themselves as the inverses are just reflections across the line y=x over the appropriate restricted domain. You need to know your trigonometric identities to solve these problems.

Example 1. Find the derivative of arcsin x.

Solution 1. We use implicit differentiation on the inverse function.

$y = \text{arcsin}\, x$

$\sin y = x$

$\cos y\cfrac{dy}{dx} = 1$

$\cfrac{dy}{dx} = \cfrac{1}{\cos y}$

$\cfrac{dy}{dx} = \cfrac{1}{\pm \sqrt{1-\sin ^{2} y} }$

$\cfrac{dy}{dx} = \cfrac{1}{\sqrt{1-x^2}}$

We have taken the positive root according to the slope of the function.

Example 2. Find the derivative of arctan x.

Solution 2. We use implicit differentiation on the inverse function.

$y = \text{arctan}\, x$

$\tan y = x$

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

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

$\cfrac{dy}{dx} = \cfrac{1}{1+\tan ^{2} y } =\cfrac{1}{1+x^{2} }$

Example 3. Find the derivative of arcsec x.

Solution 3. We use implicit differentiation on the inverse function.

$y = \text{arcsec}\, x \\\sec y = x \\\sec y \tan y \cfrac{dy}{dx} = 1 \\ \cfrac{dy}{dx} = \cfrac{1}{\sec y \tan y}\\\cfrac{dy}{dx} = \cfrac{1}{\pm \sec y\sqrt{\sec ^{2} y-1} } = \cfrac{1}{\pm x\sqrt{x^{2} -1}}\ \cfrac{dy}{dx} = \cfrac{1}{\left| x \right| \sqrt{x^2-1}}$

To replace the plus or minus sign with an absolute value we have looked at the form of the slope of the graph.

Example 4. Find the other three derivatives of inverse trigonometric functions using relationships between the angles of right triangles.

Solution 4. The two angles beside the right angle of a right triangle add up to ninety degrees. The sum of these angles can be expressed in terms of the sum of inverse trigonometric functions.

$\text{arcsin}\, x+\text{arccos}\, x =\cfrac{\pi }{2} \\\text{arctan}\, x+\text{arccot}\, x = \cfrac{\pi }{2} \\\text{arcsec}\, x+\text{arccsc}\, x = \cfrac{\pi }{2}$

By implicit differentiation of these equations we find that

$\cfrac{d}{dx} \text{arccos}\, x =-\cfrac{d}{dx} \text{arcsin}\, x \\\cfrac{d}{dx} \text{arccot}\, x =-\cfrac{d}{dx} \text{arctan}\, x \\\cfrac{d}{dx} \text{arccsc}\, x=-\cfrac{d}{dx} \text{arcsec}\, x$