Higher Secondary Mathematics: 21.10 - Derivatives of Inverse Trigonometric Functions

In the previous section, we completed a discussion on derivatives of implicit functions. In this section, we will see derivatives of inverse trigonometric functions.

• We have seen inverse trigonometric functions in chapter 18. Now we want to find the derivatives of those functions.
• We know that, only continuous functions are differentiable. Inverse trigonometric functions are continuous functions. We will see the proof in higher classes.
• At present we will see the method for differentiating those functions.

Let us see an example. It can be written in 8 steps:
1. Given that, f(x) = sin⁻¹x. We want to find f'(x).
2. Let y = sin⁻¹x. Then x = sin y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\sin y)$
⇒ $1~=~ \cos y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{1}{\cos y}~=~\frac{1}{\cos(\sin^{-1} x)}$
4. In the above result, the denominator cos y should not be equal to zero.
• That means, y should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, sin⁻¹x should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, x should not be equal to −1 or 1.
5. In chapter 18, we saw that, the acceptable domain for the sin⁻¹ function is [−1,1] (details here).
• But from the above step (4), we see that, derivative of the sin⁻¹ function is not defined at −1 and 1.
• So we can write two important points:
(i) For the sin⁻¹ function, the input can be taken from the interval [−1,1].
(ii) Derivative of the sin⁻¹ function is available only in the interval (−1,1).
6. We can eliminate the trigonometric ratios from the result.
• We have:
$\cos^2 y ~=~ 1 - \sin^2 y ~=~ 1 - [\sin(\sin^{-1} x)]^2 ~=~1 - x^2 $
So $\cos y ~=~ \pm \sqrt{1 - x^2}$
7. We have to determine whether cos y is $\sqrt{1 - x^2}$ or $- \sqrt{1 - x^2}$.
• The input for cos y is "y", which is "sin⁻¹x".
• We saw that, sin⁻¹x should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, input for cos y should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• Between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, cosine is +ve.
• So cos y is +ve.
• Thus $\cos y ~=~ \sqrt{1 - x^2}$
8. So we can write:
If y = sin⁻¹x, then $\frac{dy}{dx}~=~\frac{1}{\cos y}~=~\frac{1}{\cos(\sin^{-1} x)}~=~ \frac{1}{\sqrt{1 - x^2}}$

Now we will see the derivative of inverse cosine function. It can be written in 8 steps:

1. Given that, f(x) = cos⁻¹x. We want to find f'(x).
2. Let y = cos⁻¹x. Then x = cos y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\cos y)$
⇒ $1~=~ -\sin y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{-1}{\sin y}~=~\frac{-1}{\sin(\cos^{-1} x)}$
4. In the above result, the denominator sin y should not be equal to zero.
• That means, y should not be equal to 0 or $\pi$.
• That means, cos⁻¹x should not be equal to 0 or $\pi$.
• That means, x should not be equal to 1 or −1.
5. In chapter 18, we saw that, the acceptable domain for the cos⁻¹ function is [−1,1] (details here).
• But from the above step (4), we see that, derivative of the cos⁻¹ function is not defined at −1 and 1.
• So we can write two important points:
(i) For the cos⁻¹ function, the input can be taken from the interval [−1,1].
(ii) Derivative of the cos⁻¹ function is available only in the interval (−1,1).
6. We can eliminate the trigonometric ratios from the result.
• We have:
$\sin^2 y ~=~ 1 - \cos^2 y ~=~ 1 - [\cos(\cos^{-1} x)]^2 ~=~1 - x^2 $
So $\sin y ~=~ \pm \sqrt{1 - x^2}$
7. We have to determine whether sin y is $\sqrt{1 - x^2}$ or $- \sqrt{1 - x^2}$.
• The input for sin y is "y", which is "cos⁻¹x".
• We saw that, cos⁻¹x should not be equal to 0 or π.
• That means, input for sin y should not be equal to 0 or π.
• Between 0 and π, sine is +ve.
• So sin y is +ve.
• Thus $\sin y ~=~ \sqrt{1 - x^2}$
8. So we can write:
If y = cos⁻¹x, then $\frac{dy}{dx}~=~\frac{-1}{\sin y}~=~\frac{-1}{\sin(\cos^{-1} x)}~=~\frac{-1}{\sqrt{1 - x^2}}$

Now we will see the derivative of inverse tangent function. It can be written in 4 steps:

1. Given that, f(x) = tan⁻¹x. We want to find f'(x).
2. Let y = tan⁻¹x. Then x = tan y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\tan y)$
⇒ $1~=~ \sec^2 y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{1}{\sec^2 y}~=~\frac{1}{1 + \tan^2 y}~=~\frac{1}{1 + x^2 }$
4. In the above result, the denominator 1 + x² can never become zero.
• That means, x can be any real number.

• We have seen the derivatives of:
    ♦ inverse sine function
    ♦ inverse cosine function
    ♦ inverse tangent function

• There are three more inverse trigonometric functions:
    ♦ inverse cosecant function
    ♦ inverse secant function
    ♦ inverse cotangent function
        ✰ Derivatives of these three functions can be obtained by writing them in terms of sine, cosine or tangent. For that, we can use the first property of inverse trigonometric functions. Details here.

Let us see some solved examples.

Solved example 21.37
Find $\frac{dy}{dx}$ if y = 4 cos⁻¹ x − 10 tan⁻¹ x.
Solution: