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Derivative of Inverse Function

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Estimated time: 10 minutes
CBSE: Class 1

Definition: Inverse Function

If a function reverses the action of another function, it is called its inverse function. For example, if \[y = \sin^{-1} x\], then \[x = \sin y\], which means the inverse function converts a trigonometric value back into an angle.

CBSE: Class 12
Maharashtra State Board: Class 12

Formula: Derivative of Inverse Functions

Function Derivative Condition
sin⁻¹x \[\frac{1}{\sqrt{1-x^{2}}}\] |x| < 1
sin⁻¹(f(x)) \[\frac{1}{\sqrt{1-\{f\left(x\right)\}^{2}}}\frac{d}{dx}f\left(x\right)\] |f(x)| < 1
cos⁻¹x \[-\frac{1}{\sqrt{1-x^{2}}}\] x| < 1
cos⁻¹(f(x)) \[-\frac{1}{\sqrt{1-\left\{f\left(x\right)\right\}^{2}}}\frac{d}{dx}f(x)\] |f(x)| < 1
tan⁻¹x \[\left(\frac{1}{1+x^{2}}\right)\] x ∈ R
tan⁻¹(f(x)) \[\frac{1}{1+\left\{f\left(x\right)\right\}^{2}}\frac{d}{dx}f(x)\] f(x) ∈ R
cot⁻¹x \[-\left(\frac{1}{1+x^{2}}\right)\] x ∈ R
cot⁻¹(f(x)) \[-\frac{1}{1+\{f(x)\}^{2}}\frac{d}{dx}f(x)\] f(x) ∈ R
sec⁻¹x \[\frac{1}{|x|\sqrt{x^{2}-1}}\] |x| > 1
sec⁻¹(f(x)) \[\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] |f(x)| > 1
cosec⁻¹x \[-\left(\frac{1}{|x|\sqrt{x^{2}-1}}\right)\]

|x| > 1

cosec⁻¹(f(x)) \[-\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] |f(x)| > 1
CBSE: Class 12

Example 1

Find the derivative of \[f\] given by \[f(x) = \sin^{-1} x\] assuming it exists.

Solution: Let \[y = \sin^{-1} x\]. Then, \[x = \sin y\].

Differentiating both sides w.r.t. \[x\], we get

\[ 1 = \cos y \frac{dy}{dx} \]

which implies that 

\[ \frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\cos(\sin^{-1} x)} \]

Now use the identity:

\[\sin^2 y + \cos^2 y = 1\]

Since, sin y = x

we get

\[\cos^2 y = 1 - \sin^2 y\]
\[\cos^2 y = 1 - x^2\]

Taking square root,

\[\cos y = \sqrt{1 - x^2}\]

because for

\[y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\]
cos y is positive.

Hence,

\[\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}\]

Therefore,

\[\boxed{\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}}\] for −1 < x < 1
CBSE: Class 12
Maharashtra State Board: Class 12

Key Points: Derivative of Inverse Functions

  • The derivative of an inverse function is usually found using implicit differentiation.

  • For \[\sin^{-1} x\] and \[\cos^{-1} x\], the denominator is \[\sqrt{1 - x^2}\].

  • For \[\tan^{-1} x\] and \[\cot^{-1} x\], the denominator is \[1 + x^2\].

  • For \[\sec^{-1} x\] and \[\csc^{-1} x\], the denominator involves \[|x|\sqrt{x^2 - 1}\].

  • Negative signs are especially important in \[\cos^{-1} x\], \[\cot^{-1} x\], and \[\csc^{-1} x\].

  • Domain restrictions must be checked before applying formulas.

Shaalaa.com | Derivative Rules for Inverse Trigonometric Functions Derived

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