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Derivative of Inverse Function

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Estimated time: 5 minutes
Maharashtra State Board: Class 1

Key Points: Derivative of Inverse Functions

If y = f(x) is a differentiable function of x such that the inverse function x = f⁻¹(y) exists, then x is a differentiable function of y and

\[\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}\], where \[\frac{\mathrm{d}y}{\mathrm{d}x}\neq0\].

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

Shaalaa.com | Derivative Rules for Inverse Trigonometric Functions Derived

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