Inverse trigonometric, exponential and logarithmic functions

What you will be able to do

• write the derivatives of $e^x$, $a^x$, $\ln x$ and $\log_a x$;
• derive and use the derivatives of the inverse trigonometric functions;
• simplify an inverse trigonometric expression before differentiating.

Exponential functions

Look at the graph of $y = a^x$. Its tangent at any point has slope equal to a constant times the height. The constant is $\ln a$. So

There is one base for which the constant is $1$. It is the number $e \approx 2.718$. For this base the slope equals the height at every point:

With the chain rule, $\dfrac{d}{dx}\,e^{f(x)} = e^{f(x)}f'(x)$. For example, $\dfrac{d}{dx}\,e^{\sin x} = e^{\sin x}\cos x$.

Logarithmic functions

The second follows from $\log_a x = \dfrac{\ln x}{\ln a}$. With the chain rule, $\dfrac{d}{dx}\ln f(x) = \dfrac{f'(x)}{f(x)}$. For example, $\dfrac{d}{dx}\ln(x^2 + 1) = \dfrac{2x}{x^2 + 1}$. The derivative of $\ln|x|$ is also $\dfrac{1}{x}$, for every $x \ne 0$.

Inverse trigonometric functions

To differentiate $y = \sin^{-1}x$, write $\sin y = x$. Differentiate both sides with respect to $x$:

For $-1 < x < 1$ the angle $y$ lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, where $\cos y > 0$. So $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ and $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}$.

The same idea works for any inverse function. If $y = f^{-1}(x)$, then $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$, wherever $f'(y) \ne 0$. The graphs show why: a function and its inverse are mirror images, so their slopes at matching points are reciprocals.

The standard results are:

• $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1 - x^2}}$ and $\dfrac{d}{dx}\cos^{-1}x = -\dfrac{1}{\sqrt{1 - x^2}}$, for $|x| < 1$;
• $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1 + x^2}$ and $\dfrac{d}{dx}\cot^{-1}x = -\dfrac{1}{1 + x^2}$;
• $\dfrac{d}{dx}\sec^{-1}x = \dfrac{1}{|x|\sqrt{x^2 - 1}}$ and $\dfrac{d}{dx}\operatorname{cosec}^{-1}x = -\dfrac{1}{|x|\sqrt{x^2 - 1}}$, for $|x| > 1$.

A-Level Further Maths, IB Higher Level and AP Calculus use arcsin, arccos and arctan for $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$. The derivatives of the inverse secant and cosecant are asked mainly in Indian exams.

Simplify first

An expression such as $\tan^{-1}\dfrac{2x}{1 - x^2}$ is messy to differentiate. A substitution $x = \tan\theta$ turns it into $\tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1}x$, for $|x| < 1$. Then the derivative is $\dfrac{2}{1 + x^2}$. Always check that the angle lies in the range of the inverse function before you cancel.

A worked example

Differentiate $y = \tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x}$.

1. Substitute. Put $x = \tan\theta$ with $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$. Then $\sqrt{1 + x^2} = \sec\theta$.
2. Simplify. $\dfrac{\sec\theta - 1}{\tan\theta} = \dfrac{1 - \cos\theta}{\sin\theta} = \tan\dfrac{\theta}{2}$.
3. Remove the inverse. $y = \tan^{-1}\left(\tan\dfrac{\theta}{2}\right) = \dfrac{\theta}{2} = \dfrac{1}{2}\tan^{-1}x$.
4. Differentiate. $\dfrac{dy}{dx} = \dfrac{1}{2(1 + x^2)}$.

Practise with new numbers

Try it in Python

The program checks six standard derivatives. For each one it prints the value from the formula and a numerical slope at $x = 0.5$.

Common mistakes

• Mixing $\sin^{-1}x$ with $\dfrac{1}{\sin x}$. The symbol $\sin^{-1}x$ means the inverse function, not the reciprocal.
• Losing the chain factor. The derivative of $\tan^{-1}(3x)$ is $\dfrac{3}{1 + 9x^2}$, not $\dfrac{1}{1 + 9x^2}$.
• The wrong sign for cosine. The derivative of $\cos^{-1}x$ is the negative of that of $\sin^{-1}x$.
• Using the power rule on $a^x$. The derivative of $a^x$ is $a^x\ln a$, not $xa^{x-1}$.

Remember these

• $(e^x)' = e^x$ and $(a^x)' = a^x\ln a$.
• $(\ln x)' = \dfrac{1}{x}$ and $(\log_a x)' = \dfrac{1}{x\ln a}$.
• $(\sin^{-1}x)' = \dfrac{1}{\sqrt{1 - x^2}}$, $(\cos^{-1}x)' = -\dfrac{1}{\sqrt{1 - x^2}}$, $(\tan^{-1}x)' = \dfrac{1}{1 + x^2}$.
• If $y = f^{-1}(x)$, then $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Differentiate $y = e^{3x}$.

2. Differentiate $y = \ln(\sin x)$.

3. Differentiate $y = 5^x$.

4. Differentiate $y = \sin^{-1}(2x)$.

5. Show that $\dfrac{d}{dx}\left(\sin^{-1}x + \cos^{-1}x\right) = 0$, and say why.

References

• OpenStax, Calculus Volume 1, Chapter 3, sections on derivatives of inverse functions and of exponential and logarithmic functions.
• NCERT, Mathematics Part I for Class XII, Chapter 5, sections on inverse trigonometric, exponential and logarithmic functions.