  ## Derivatives of Inverse Trigonometric Functions

We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest:

Finding the Derivative of Inverse Sine Function, $\displaystyle{\frac{d}{dx} (\arcsin x)}$

Suppose $\arcsin x = \theta$. Then it must be the cases that

$$\sin \theta = x$$

Implicitly differentiating the above with respect to $x$ yields

$$(\cos \theta) \cdot \frac{d\theta}{dx} = 1$$

Dividing both sides by $\cos \theta$ immediately leads to a formula for the derivative.

$$\frac{d\theta}{dx} = \frac{1}{\cos \theta}$$

To be a useful formula for the derivative of $\arcsin x$ however, we would prefer that $\displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arcsin x)}$ be expressed in terms of $x$, not $\theta$.

Upon considering how to then replace the above $\cos \theta$ with some expression in $x$, recall the pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$ and what this identity implies given that $\sin \theta = x$:

$$\cos^2 \theta = 1 - x^2$$

So we know either $\cos \theta$ is then either the positive or negative square root of the right side of the above equation. Since $\theta$ must be in the range of $\arcsin x$ (i.e., $[-\pi/2,\pi/2]$), we know $\cos \theta$ must be positive. Thus,

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

Finally, plugging this into our formula for the derivative of $\arcsin x$, we find

$$\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}$$

Finding the Derivative of Inverse Cosine Function, $\displaystyle{\frac{d}{dx} (\arccos x)}$

The process for finding the derivative of $\arccos x$ is almost identical to that used for $\arcsin x$:

Suppose $\arccos x = \theta$. Then it must be the case that

$$\cos \theta = x$$

Implicitly differentiating the above with respect to $x$ yields

$$(-\sin \theta) \cdot \frac{d\theta}{dx} = 1$$

Dividing both sides by $-\sin \theta$ immediately leads to a formula for the derivative.

$$\frac{d\theta}{dx} = \frac{-1}{\sin \theta}$$

To be a useful formula for the derivative of $\arccos x$ however, we would prefer that $\displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arccos x)}$ be expressed in terms of $x$, not $\theta$.

Upon considering how to then replace the above $\sin \theta$ with some expression in $x$, recall the pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$ and what this identity implies given that $\cos \theta = x$:

$$\sin^2 \theta = 1 - x^2$$

So we know either $\sin \theta$ is then either the positive or negative square root of the right side of the above equation. Since $\theta$ must be in the range of $\arccos x$ (i.e., $[0,\pi]$), we know $\sin \theta$ must be positive. Thus,

$$\sin \theta = \sqrt{1-x^2}$$

Finally, plugging this into our formula for the derivative of $\arccos x$, we find

$$\frac{d}{dx} (\arccos x) = \frac{-1}{\sqrt{1-x^2}}$$

Finding the Derivative of the Inverse Tangent Function, $\displaystyle{\frac{d}{dx} (\arctan x)}$

The process for finding the derivative of $\arctan x$ is slightly different, but the same overall strategy is used:

Suppose $\arctan x = \theta$. Then it must be the case that

$$\tan \theta = x$$

Implicitly differentiating the above with respect to $x$ yields

$$(\sec^2 \theta) \cdot \frac{d\theta}{dx} = 1$$

Dividing both sides by $\sec^2 \theta$ immediately leads to a formula for the derivative.

$$\frac{d\theta}{dx} = \frac{-1}{\sec^2 \theta}$$

To be a useful formula for the derivative of $\arctan x$ however, we would prefer that $\displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arctan x)}$ be expressed in terms of $x$, not $\theta$.

Upon considering how to then replace the above $\sec^2 \theta$ with some expression in $x$, recall the other pythagorean identity $\tan^2 \theta + 1 = \sec^2 \theta$ and what this identity implies given that $\tan \theta = x$:

$$\sec^2 \theta = 1 + x^2$$

Not having to worry about the sign, as we did in the previous two arguments, we simply plug this into our formula for the derivative of $\arccos x$, to find

$$\frac{d}{dx} (\arctan x) = \frac{1}{1 + x^2}$$

Finding the Derivative of the Inverse Cotangent Function, $\displaystyle{\frac{d}{dx} (\textrm{arccot } x)}$

The derivative of $\textrm{arccot } x$ can be found similarly. Suppose $\textrm{arccot } x = \theta$. Then $\cot \theta = x$. Implicitly differentiating with respect to $x$ yields $$-csc^2 \theta \cdot \frac{d\theta}{dx} = 1$$ which implies the following, upon realizing that $\cot \theta = x$ and the identity $\cot^2 \theta + 1 = \csc^2 \theta$ requires $\csc^2 \theta = 1 + x^2$, $$\frac{d\theta}{dx} = \frac{-1}{\csc^2 \theta} = \frac{-1}{1+x^2}$$ Thus, $$\frac{d}{dx}(\textrm{arccot } x) = \frac{-1}{1+x^2}$$

Finding the Derivative of the Inverse Secant Function, $\displaystyle{\frac{d}{dx} (\textrm{arcsec } x)}$

The derivative of $\textrm{arcsec } x$ is more interesting...

Here, we suppose $\textrm{arcsec } x = \theta$, which means $sec \theta = x$. Like before, we differentiate this implicitly with respect to $x$ to find

$$\sec \theta \tan \theta \frac{d\theta}{dx} = 1$$

Solving for $d\theta/dx$ in terms of $\theta$ we quickly get

$$\frac{d\theta}{dx} = \frac{1}{\sec \theta \tan \theta}$$

This is where we need to be careful. Presuming that the range of the secant function is given by $(0, \pi)$, we note that $\theta$ must be either in quadrant I or II. In both, the product of $\sec \theta \tan \theta$ must be positive. This implies

$$\sec \theta \tan \theta = |\sec \theta||\tan \theta|$$

Of course $|\sec \theta| = |x|$, and we can use $\tan^2 \theta + 1 = \sec^2 \theta$ to establish $|\tan \theta| = \sqrt{x^2 - 1}$. As such,

$$\frac{d}{dx}(\textrm{arcsec } x) = \frac{1}{|x|\sqrt{x^2 - 1}}$$