Derivatives of inverse trigonometric functions | Class 12 Maths
Derivative of Inverse Sine Function
$f(x)= sin^{–1} x$
Let $y = sin^{–1} x$
siny =x
Differentiating both sides w.r.t. x, we get
$cos y \frac {dy}{dx}=1$
or
$\frac {dy}{dx} = \frac {1}{cos y} = \frac {1}{cos sin^{–1} x}$
Clearly we can see that x cannot be -1 or 1 as then it will then become undefined. Therefore $x \in (-1,1)$
Also
$cos^2 y=1 -sin^2 y$
$cos^2 y = 1 -x^2$
As $sin^{–1} x$ lies between $(-\pi/2, \pi/2)$, cos y is always positive
Hence
$cos y = \sqrt {1-x^2}$
Therefore
$\frac {dy}{dx} = \frac {1}{\sqrt {1-x^2}}$
Domain of f' is (-1,1)
Derivative of Inverse cos Function
$f(x)= cos^{–1} x$
Let $y = cos^{–1} x$
cosy =x
Differentiating both sides w.r.t. x, we get
$-sin y \frac {dy}{dx}=1$
or
$\frac {dy}{dx} = -\frac {1}{sin y} = -\frac {1}{sin cos^{–1} x}$
Clearly we can see that x cannot be -1 or 1 as then it will then become undefined. Therefore $x \in (-1,1)$
Also
$sin^2 y=1 -cos^2 y$
$sin^2 y = 1 -x^2$
As $cos^{–1} x$ lies between $(0, \pi)$, sin y is always positive
Hence
$sin y = \sqrt {1-x^2}$
Therefore
$\frac {dy}{dx} = -\frac {1}{\sqrt {1-x^2}}$
Domain of f' is (-1,1)
Derivative of Inverse tan Function
$f(x)= tan^{–1} x$
Let $y = tan^{–1} x$
tany =x
Differentiating both sides w.r.t. x, we get
$sec^2 y \frac {dy}{dx}=1$
or
$\frac {dy}{dx} = \frac {1}{sec^2 y} = \frac {1}{1 + tan^2 y}= \frac {1}{1 + x^2}$
Domain of f' is R
Derivative of Inverse cot Function
$f(x)= cot^{–1} x$
Let $y = cot^{–1} x$
cot y =x
Differentiating both sides w.r.t. x, we get
$-cosec^2 y \frac {dy}{dx}=1$
or
$\frac {dy}{dx} = -\frac {1}{cosec^2 y} = \frac {1}{1 + cot^2 y}= -\frac {1}{1 + x^2}=$
Domain of f' is R
Derivative of Inverse cosec Function
$f(x)= cosec^{–1} x$
Let $y = cosec^{–1} x$
cosec y =x
Differentiating both sides w.r.t. x, we get
$-cosec y cot y \frac {dy}{dx}=1$
or
$\frac {dy}{dx} =- \frac {1}{cosec y cot y} $
Clearly we can see that x cannot be -1 or 1 as then it will then become undefined. Therefore $x \in R -[-1,1]$
Also
$cot^2 y=cosec^2 y -1$
$cot^2 y =x^2 -1 $
As $cosec^{–1} x$ lies between $(-\pi/2, \pi/2)$, cot y will be positive and negative respectively
Hence
When x is negative
$cot y = \sqrt {x^2 -1}$
When x is psotive
$cot y =- \sqrt {x^2 -1}$
Therefore
When x is negative
$\frac {dy}{dx} = -\frac {1}{- x \sqrt {x^2 -1}}=-\frac {1}{ |x| \sqrt {x^2 -1}} $
When is positive
$\frac {dy}{dx} = -\frac {1}{ x \sqrt {x^2 -1}}=-\frac {1}{ |x| \sqrt {x^2 -1}} $
Domain of f' is R -[-1,1]
Derivative of Inverse sec Function
$f(x)= sec^{–1} x$
Let $y = sec^{–1} x$
sec y =x
Differentiating both sides w.r.t. x, we get
$sec y tan y \frac {dy}{dx}=1$
or
$\frac {dy}{dx} =\frac {1}{sec y tan y} $
Clearly we can see that x cannot be -1 or 1 as then it will then become undefined. Therefore $x \in R -[-1,1]$
Also
$tan^2 y=sec^2 y -1$
$tan^2 y =x^2 -1 $
As $sec^{–1} x$ lies between $(0, \pi)$, tan y will be positive and negative respectively
Hence
$tan y =\pm \sqrt {x^2 -1}$
Therefore
$\frac {dy}{dx} = \frac {1}{\pm x \sqrt {x^2 -1}} =\frac {1}{ |x| \sqrt {x^2 -1}} $
Domain of f' is R -[-1,1]
Summary
Solved examples
Question 1
Find the derivative of \( y = cos^{-1}(3x) \). Solution
First, recall the derivative of \( \arccos(x) \), which is \( -\frac{1}{\sqrt{1 - x^2}} \).
Then, apply the chain rule:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - (3x)^2}} \cdot \frac{d}{dx}(3x) \]
\[ = -\frac{1}{\sqrt{1 - 9x^2}} \cdot 3 \]
\[ = -\frac{3}{\sqrt{1 - 9x^2}} \]
Question 2
Find the derivative of \( y = x \cdot \arctan(x) \). Solution
Here, we need to use the product rule, which is \( (uv)' = u'v + uv' \), where \( u = x \) and \( v = tan^{-1}(x) \).
Now The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1 + x^2} \).
So,
\[ \frac{dy}{dx} = 1 \cdot \arctan(x) + x \cdot \frac{1}{1 + x^2} \]
\[ = \arctan(x) + \frac{x}{1 + x^2} \]
Question 3
Find the derivative of \( y = \frac{\sin^{-1}(x)}{x} \) for \( x \neq 0 \). Solution
Apply the quotient rule, \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = \sin^{-1}(x) \) and \( v = x \).
Now The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \).
So,
\[ \frac{dy}{dx} = \frac{\frac{1}{\sqrt{1 - x^2}} \cdot x - \sin^{-1}(x) \cdot 1}{x^2} \]
\[ = \frac{x}{x^2\sqrt{1 - x^2}} - \frac{\sin^{-1}(x)}{x^2} \]
\[ = \frac{1}{x\sqrt{1 - x^2}} - \frac{\sin^{-1}(x)}{x^2} \]
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