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⁻¹ x, then $\frac{dy}{dx}~=~\frac{1}{\cos y}~=~\frac{1}{\cos(\sin^{-1} x)}~=~ \frac{1}{\sqrt{1 - x^2}}$

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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⁻¹ x, then $\frac{dy}{dx}~=~\frac{-1}{\sin y}~=~\frac{-1}{\sin(\cos^{-1} x)}~=~\frac{-1}{\sqrt{1 - x^2}}$

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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.

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• 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.

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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:

Solved example 21.38
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, \sqrt{x} \sin^{-1} x$.
Solution:

 

Solved example 21.39
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, \frac{1}{\sin^{-1} x}$
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{d}{dx}\left(\sin^{-1} x \right)^{-1}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{-1 \left(\sin^{-1} x \right)^{-2} \frac{d}{dx}\left(\sin^{-1} x \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{\left(\sin^{-1} x \right)^{2} \sqrt{1 – x^2}}}    \\
\end{array}$

Solved example 21.40
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, x \tan^{-1} \sqrt{x}$
Solution:

 

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Link to a few more solved examples is given below:

Exercise 21.3

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We have completed a discussion on the derivatives of inverse trigonometric functions. In the next section, we will see Exponential functions.

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21.9 - Derivatives of Implicit Functions - Easy method

In the previous section, we saw the method to find the derivatives of implicit functions. In this section, we will see some solved examples. In those solved examples, we will see an easy method also.

Solved example 21.32
Find $\frac{dy}{dx}$ if x + sin (xy) − y = 0.
Solution:
1. Consider the function x + sin (xy) − y = 0.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
x + sin (x f(x)) − f(x) = 0
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) \,+\, \frac{d}{dx} \sin(x f(x)) \, - \, \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (0)$
⇒ $1 \,+\, \frac{d}{dx} \sin(x f(x)) \, - \, f'(x) \,=\, 0$
4. The second term in the L.H.S needs special attention. It is the derivative of a composite function. It can be determined in 4 steps:
(i) g(x) = sin (x f(x)) = (v∘u)(x) = v(u(x))
Where u(x) = x f(x) and v(u(x)) = sin (x f(x)).
(ii) v'(u(x)) = cos (x f(x))
(iii) u'(x) = 1.f(x) + x f'(x)
(iv) So g'(x) = v'(u(x)).u'(x)
= f(x).cos (x f(x)) + x f'(x) cos(x f(x))
5. So the result in (3) becomes:
$1 \,+\, f(x).\cos (x f(x)) + x. f'(x). \cos(x f(x)) \, - \, f'(x) \,=\, 0$
⇒ $1 \,+\, f(x).\cos (x f(x)) ~+~  f'(x) \left[x. \cos(x. f(x)) \, - \, 1 \right] \,=\, 0$
⇒ $1 \,+\, f(x).\cos (x f(x)) ~-~  f'(x) \left[1~-~x. \cos(x. f(x))  \right] \,=\, 0$
⇒ $f'(x) = \frac{1 \,+\, f(x).\cos (x. f(x))}{1~-~x. \cos(x. f(x))}$
6. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx} = \frac{1 \,+\, y \cos (xy)}{1~-~x \cos(xy)}$

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Once we become familiar with the basics of this method,
    ♦ we need not write "f(x)" in the place of "y". 
        ✰ we can keep "y" as such.
    ♦ we need not write "f'(x)" to show differentiation of y.
        ✰ we can write $\frac{dy}{dx}$ instead.
    ♦ Also in simple cases, we may be able to apply chain rule in just one step.

So let us redo the above solved example.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x)\,+\,\frac{d}{dx} (\sin(xy))\,-\,\frac{d}{dx} (y)}    & {~=~}    &{\frac{d}{dx} (0)}    \\
{~\color{magenta}    2    }    &{\implies}    &{1\,+\,\frac{d}{dx} (\sin(xy))\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{1\,+\,\cos(xy) \left[\frac{d}{dx} (x) .y + x.\frac{dy}{dx} \right]\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{1\,+\,\cos(xy) \left[y + x.\frac{dy}{dx} \right]\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{1\,+\,y \cos(xy) + x.\frac{dy}{dx} \cos(xy)\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{\implies}    &{1\,+\,y \cos(xy) + \frac{dy}{dx} \left[x \cos(xy)\,-\,1 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{\implies}    &{1\,+\,y \cos(xy)}    & {~=~}    &{\frac{dy}{dx} \left[1 - x \cos(xy) \right]}    \\
{~\color{magenta}    8    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1\,+\,y \cos(xy)}{1 - x \cos(xy)}}    \\
\end{array}$

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Solved example 21.33
Find $\frac{dy}{dx}$ if x³y⁵ + 3x = 8y³ + 1.
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x^3 y^5)\,+\,\frac{d}{dx} (3x)}    & {~=~}    &{\frac{d}{dx} (8 y^3)\,+\,\frac{d}{dx} (1)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{d}{dx} (x^3 y^5)\,+\,3}    & {~=~}    &{\frac{d}{dx} (8 y^3)\,+\,0}    \\
{~\color{magenta}    3    }    &{\implies}    &{3 x^2 . y^5 \,+\,x^3 . \frac{d}{dx} (y^5) \,+\,3}    & {~=~}    &{8 \frac{d}{dx} (y^3)}    \\
{~\color{magenta}    4    }    &{\implies}    &{3 x^2 . y^5 \,+\,x^3 . \left[5 y^4 .  \frac{dy}{dx} \right] \,+\,3}    & {~=~}    &{8. 3 y^2 . \frac{dy}{dx}}    \\
{~\color{magenta}    5    }    &{\implies}    &{3 x^2 y^5 \,+\, 5 x^3 y^4 \frac{dy}{dx} \,+\, 3}    & {~=~}    &{24 y^2 \frac{dy}{dx}}    \\
{~\color{magenta}    6    }    &{\implies}    &{3 x^2 y^5  \,+\, 3}    & {~=~}    &{24 y^2 \frac{dy}{dx} \,-\, 5 x^3 y^4 \frac{dy}{dx}}    \\
{~\color{magenta}    7    }    &{\implies}    &{3 x^2 y^5  \,+\, 3}    & {~=~}    &{\left[24 y^2  \,-\, 5 x^3 y^4 \right]\frac{dy}{dx}}    \\
{~\color{magenta}    8    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{3 x^2 y^5  \,+\, 3}{24 y^2  \,-\, 5 x^3 y^4}}    \\
\end{array}$

◼ Remarks:
• 3 (magenta color). For differentiating x³y⁵, we use product rule. For differentiating y⁵, we treat it as a composite function.

• In the R.H.S, y³ is also treated as a composite function.

Solved example 21.34
Find $\frac{dy}{dx}$ if x² + xy + y² = 100.
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x^2)\,+\,\frac{d}{dx} (xy) \,+\,\frac{d}{dx} (y^2)}    & {~=~}    &{\frac{d}{dx} (100)}    \\
{~\color{magenta}    2    }    &{\implies}    &{2x\,+\,\frac{d}{dx} (xy) \,+\,\frac{d}{dx} (y^2)}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{2x\,+\,\left[1. {y} + x \frac{dy}{dx} \right] \,+\,\left[2y \frac{dy}{dx} \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{2x + y + (x+2y)\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{2x + y}    & {~=~}    &{- (x+2y)\frac{dy}{dx}}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{-(2x + y)}{(x+2y)}}    \\
\end{array}$

Solved example 21.35
Find $\frac{dy}{dx}$ if x³ + x²y + xy² + y³ = 81.
Solution:

Solved example 21.36
Find $\frac{dy}{dx}$ if sin²y + cos xy = π.
Solution:

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We have completed a discussion on the derivatives of implicit functions. In the next section, we will see derivatives of inverse trigonometric functions.

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21.8 - Derivatives of Implicit Functions

In the previous section, we completed a discussion on chain rule. We saw some solved examples also. In this section, we will see derivatives of implicit functions.

First we will see some basic details about explicit functions. It can be written in 7 steps:
1. Consider the equation: x−y−π = 0
2. This equation has two variables x and y. The equation can be rearranged and written as:
y = x − π
After the rearrangement, the equation has become a function. We see that, y is a function of x.
3. We can write either of the two ways:
    ♦ y = x − π
    ♦ f(x) = x − π
Now it is easy to find the derivative $\frac{dy}{dx}$ or f'(x).
4. So there are functions which can be easily rearranged so that three conditions are satisfied:
(i) There is a single "y" variable on the L.H.S.
(ii) All "x terms" are on the R.H.S.
(iii) There is no "y" variable or "y term" on the R.H.S.
• Such functions are called explicit functions.
5. The equation xy = 1 is another example.
• We can rearrange this equation and write: $y=\frac{1}{x}$
• All three conditions that we wrote in (4) are satisfied.
• So xy = 1 is an explicit function.
6.In the case of explicit functions, We say that:
y is given as an explicit function of x
7. The dictionary meaning of 'explicit' is:
Clear and exact. There will not be any need for doubts.
• Indeed, in the case of explicit functions, we can clearly see how 'y' is dependent on 'x'.

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Now we will see some basic details about implicit functions. It can be written in 5 steps:
1. Consider the equation: x+sin xy −y = 0
2. This equation has two variables x and y. The equation cannot be easily rearranged so as to satisfy the three conditions that we wrote in the case of explicit functions.
• Such functions are called implicit functions.
3. In the case of implicit functions also, "y" is dependent on "x". But the manner of this dependency is not clear and exact.
• We say that:
y is given as an implicit function of x.
That is., dependency of 'y' on 'x' is implied.
4. The dictionary meaning of 'implicit' is:
A fact is implied, but not communicated directly.
5. The equation x³y⁵ + 3x = 8y³ + 1
is another example.
• We cannot rearrange it easily.

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Let us see how to find the derivative in the case of implicit functions. It can be explained in steps:

So far in this chapter, we have been finding the derivatives of explicit functions.
Consider the explicit function f(x) = 2x+3.
• This can be written as y = 2x+3 also.
• Using 'y' instead of 'f(x)' is useful when we plot the graph of the function.
   ♦ We plot the input values along the x-axis.
   ♦ The output values are 'f(x) values'. They are plotted along the y-axis.
   ♦ So using 'y' instead of 'f(x)' will not make any difference.

• We will now see a new method for finding the derivative. It can be explained using an example.
Let us first use an explicit function as example. The method can be written in 5 steps:
1. Consider the function x−y = π.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
x−f(x) = π
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) \,-\, \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (\pi)$
⇒ 1 − f'(x) = 0
⇒ f'(x) = 1
4. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx}$ = 1 
5. We will get the same result even if we begin with y = x−π.

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Let us see another example to demonstrate the new method. This time also, we will use an explicit function.

1. Consider the function xy = 1.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
x f(x) = 1
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) . f(x) \,+\, x \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (1)$
⇒ 1.f(x) + x f'(x) = 0
⇒ f'(x) = $-\frac{f(x)}{x}$
4. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx} = -\frac{f(x)}{x} = - \frac{y}{x} = - \frac{1/x}{x} = -\frac{1}{x^2}$
5. We will get the same result even if we begin with $y=\frac{1}{x}$.

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Now we will apply the new method to an implicit function.

1. Consider the function y + sin y = cos x.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
f(x) + sin (f(x)) = cos x
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (f(x)) \,+\, \frac{d}{dx} \sin(f(x)) \,=\, \frac{d}{dx} (\cos x)$
⇒ $f'(x) \,+\, \frac{d}{dx} \sin(f(x)) \,=\, - \sin x$
4. The second term in the L.H.S needs special attention. It is the derivative of a composite function. It can be determined in 4 steps:
(i) g(x) = sin (f(x)) = (v∘u)(x) = v(u(x))
Where u(x) = f(x) and v(u(x)) = sin (f(x)).
(ii) v'(u(x)) = cos (f(x))
(iii) u'(x) = f'(x)
(iv) So g'(x) = v'(u(x)).u'(x) = cos (f(x)).f'(x)
5. So the result in (3) becomes:
$f'(x) \,+\,  \cos(f(x)).f'(x) \,=\, - \sin x$
⇒ $f'(x) \left[1 \,+\,  \cos(f(x)) \right] \,=\, - \sin x$
⇒ $f'(x) \,=\, - \frac{\sin x}{1 \,+\,  \cos(f(x))}$
6. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx} = - \frac{\sin x}{1 \,+\,  \cos(f(x))} = - \frac{\sin x}{1 \,+\,  \cos y}$
• Here, the denominator should not be zero. That means, cos y should not be −1.
• So we can write: y ≠ (2n+1)π

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In the next section, we will see some solved examples.

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