Derivatives of inverse-trigonometric functions

There are 3 versions: \begin{eqnarray} \sin^2 \theta + \cos^2 \theta &=& 1, \\ 1 + \cot^2 \theta &=& \csc^2 \theta, \text{ and} \\ \tan^2 \theta + 1 &=& \sec^2 \theta. \end{eqnarray} They are easy to derive, so don't bother memorizing them all (just the first one is enough).


Derivations:

1. The basic one is $$\sin^2 \theta + \cos^2 \theta = 1.$$ From this, we can divide both sides by either $\sin^2 \theta$ or $\cos^2 \theta$ to get the other two versions.

2. Divide by $\sin^2 \theta$: $$ \dfrac{\sin^2 \theta}{\sin^2 \theta} + \dfrac{\cos^2 \theta}{\sin^2 \theta} = \dfrac{1}{\sin^2 \theta}, $$ which simplifies to $$1 + \cot^2 \theta = \csc^2 \theta.$$
3. Divide by $\cos^2 \theta$: $$ \dfrac{\sin^2 \theta}{\cos^2 \theta} + \dfrac{\cos^2 \theta}{\cos^2 \theta} = \dfrac{1}{\cos^2 \theta}, $$ which simplifies to $$\tan^2 \theta + 1 = \sec^2 \theta.$$

Trick: For inverse trig problem of the type $y = f^{-1}(x)$, we follow a 3-step processes.

1. Calculate the derivative of $x$ with respect to $y$. That is, calculate $$\dfrac{dx}{dy} = f'(y)$$ since $x = f(y)$.

2. Make use of $x = f(y)$ and a trigonometric version of the Pythagorean theorem to rewrite $f'(y)$ in terms of $x$.

3. Flip the derivative to get the answer: $\left(\dfrac{dy}{dx} = \dfrac{1}{dx / dy}\right)$.

These steps should become clearer after following the solution below.

Solution:
1. Let us rewrite $y = \sin^{-1} (x)$ as $$x = \sin y.$$
Its derivative is $$\dfrac{dx}{dy} = \cos y.$$

2. Making use of $x = \sin y$ (since we're told $y = \sin^{-1} (x)$) and a trig version of the Pythagorean theorem, $$\sin^2 \theta + \cos^2 \theta = 1,$$ we are able to rewrite the derivative in terms of $x$: \begin{eqnarray} \dfrac{dx}{dy} &=& \cos y \\ \ &=& \sqrt{\cos^2 y} \\ \ &=& \sqrt{1 - \sin^2 y} \\ \ &=& \sqrt{1-x^2}. \end{eqnarray}

3. We then flip that to get $$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-x^2}}.$$

We follow the 3 steps laid out in the previous solution.

Solution:
1. Let us rewrite $y = \cos^{-1} (x)$ as $$x = \cos y.$$
Its derivative is $$\dfrac{dx}{dy} = -\sin y.$$

2. Making use of $x = \cos y$ (since we're told $y = \cos^{-1} (x)$) and a trig version of the Pythagorean theorem, $$\sin^2 \theta + \cos^2 \theta = 1,$$ we are able to rewrite the derivative in terms of $x$: \begin{eqnarray} \dfrac{dx}{dy} &=& -\sin y \\ \ &=& -\sqrt{\sin^2 y} \\ \ &=& -\sqrt{1 - \cos^2 y} \\ \ &=& -\sqrt{1-x^2}. \end{eqnarray}

3. We then flip that to get $$\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1-x^2}}.$$

We follow the 3 steps laid out in a previous solution.

Solution:
1. Let us rewrite $y = \tan^{-1} (x)$ as $$x = \tan y.$$
Its derivative is $$\dfrac{dx}{dy} = \sec^2 y.$$

2. We want to make use of $x = \tan y$ and a trig version of the Pythagorean theorem, $$\tan^2 \theta + 1 = \sec^2 \theta .$$ (Note: we can derive the theorem by dividing both sides of $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2 \theta$.)

With these, we are able to rewrite the derivative in terms of $x$: \begin{eqnarray} \dfrac{dx}{dy} &=& \sec^2 y \\ \ &=& 1 + \tan^2 y \\ \ &=& 1 + x^2. \end{eqnarray}

3. We then flip that to get $$\dfrac{dy}{dx} = \dfrac{1}{1 + x^2}.$$

We follow the 3 steps laid out in a previous solution.

Solution:
1. Let us rewrite $y = \sec^{-1} (x)$ as $$x = \sec y.$$
Its derivative is $$\dfrac{dx}{dy} = \tan (y) \sec (y).$$

2. We want to make use of $x = \sec (y)$ and a trig version of the Pythagorean theorem, $$\tan^2 \theta + 1 = \sec^2 \theta .$$ (Note: we can derive the theorem by dividing both sides of $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2 \theta$.)

With these, we are able to rewrite the derivative in terms of $x$: \begin{eqnarray} \dfrac{dx}{dy} &=& \tan (y) \sec (y) \\ \ &=& \sec (y) \sqrt {\tan^2 (y)} \\ \ &=& \sec (y) \sqrt {\sec^2 (y) - 1} \\ \ &=& x \sqrt {x^2 - 1}. \end{eqnarray}

3. We then flip that to get $$\dfrac{dy}{dx} = \dfrac{1}{x \sqrt {x^2 - 1}}.$$

Since we know that the derivatives of $\sin^{-1} \theta$ and $\tan^{-1} \theta$ (see previous problems) are $$ \dfrac{d \sin^{-1} \theta}{d\theta} = \dfrac{1}{\sqrt{1-\theta^2}}$$ and $$ \dfrac{d \tan^{-1} \theta}{d\theta} = \dfrac{1}{1+\theta^2},$$ we can easily solve this using the chain rule.


\begin{eqnarray} \dfrac{dy}{dx} &=& \dfrac{d \sin^{-1}(\tan^{-1} x)}{dx} \\ \ &=& \dfrac{d \sin^{-1} (\tan^{-1} x)}{d (\tan^{-1} x)} \dfrac{d (\tan^{-1} x)}{dx} \\ \ &=& \dfrac{1}{\sqrt{1-(\tan^{-1} x)^2}}\dfrac{1}{1+x^2} \end{eqnarray}

Since we know what derivatives of $\sin^{-1} t$ and $t^{-3/2}$ with respect to $t$ are, this problem can be easily solved using the chain rule. It is \begin{eqnarray} \dfrac{dy}{dx} &=& \dfrac{d(\sin^{-1} (x))^{-3/2}}{d \sin^{-1} x} \dfrac{d\sin^{-1}x}{dx} \\ \ &=& \left[\dfrac{-3}{2}(\sin^{-1} x)^{(-3/2 - 1)} \right]\dfrac{\sin^{-1} x}{dx} \\ \ &=& \dfrac{-3}{2}(\sin^{-1} x)^{-5/2} \dfrac{1}{\sqrt{1 - x^2}}. \end{eqnarray}