Question:
If \( y=\text{sech}^{-1}\left(\frac{9}{9x^{2}+10}\right) \), then \( \frac{dy}{dx} = \)
If \( y=\text{sech}^{-1}\left(\frac{9}{9x^{2}+10}\right) \), then \( \frac{dy}{dx} = \)
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Converting \( \text{sech}^{-1}(u) \) to \( \cosh^{-1}(1/u) \) is a smart shortcut. It transforms a complex fractional argument into a clean polynomial string, making the subsequent differentiation much easier.
Updated On: Jun 8, 2026
- \( \frac{-18x}{\sqrt{(9x^{2}+10)^{2}+81}} \)
- \( \frac{-18x}{\sqrt{(9x^{2}+10)^{2}-81}} \)
- \( \frac{18x}{\sqrt{(9x^{2}+19)(9x^{2}+1)}} \)
- \( \frac{18x(9x^{2}+10)}{\sqrt{(9x^{2}+19)(9x^{2}+1)}} \)
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The Correct Option is C
Solution and Explanation
Concept:
The standard derivative of the inverse hyperbolic secant function is:
\[
\frac{d}{dx}[\text{sech}^{-1}(u)] = -\frac{1}{u\sqrt{1-u^2}} \cdot \frac{du}{dx}
\]
Alternatively, we can use the identity \( \text{sech}^{-1}(z) = \cosh^{-1}\left(\frac{1}{z}\right) \) to simplify the function before differentiating.
Step 1: Simplifying the function using properties of inverse hyperbolic functions.
Using the identity \( \text{sech}^{-1}(u) = \cosh^{-1}\left(\frac{1}{u}\right) \): \[ y = \cosh^{-1}\left(\frac{9x^{2}+10}{9}\right) = \cosh^{-1}\left(x^2 + \frac{10}{9}\right) \]
Step 2: Differentiating with respect to \( x \).
The derivative of \( \cosh^{-1}(u) \) is \( \frac{1}{\sqrt{u^2-1}} \cdot \frac{du}{dx} \). Let \( u = x^2 + \frac{10}{9} \): \[ \frac{du}{dx} = 2x \] Substituting into the formula: \[ \frac{dy}{dx} = \frac{1}{\sqrt{\left(x^2 + \frac{10}{9}\right)^2 - 1}} \cdot 2x \]
Step 3: Simplifying the algebraic denominator expression.
Expand the expression inside the square root: \[ \left(x^2 + \frac{10}{9}\right)^2 - 1 = \left(x^2 + \frac{10}{9} - 1\right)\left(x^2 + \frac{10}{9} + 1\right) = \left(x^2 + \frac{1}{9}\right)\left(x^2 + \frac{19}{9}\right) \] \[ = \frac{(9x^2+1)(9x^2+19)}{81} \] Taking the square root brings a factor of 9 to the numerator: \[ \frac{dy}{dx} = \frac{2x}{\frac{\sqrt{(9x^2+1)(9x^2+19)}}{9}} = \frac{18x}{\sqrt{(9x^2+19)(9x^2+1)}} \]
Step 1: Simplifying the function using properties of inverse hyperbolic functions.
Using the identity \( \text{sech}^{-1}(u) = \cosh^{-1}\left(\frac{1}{u}\right) \): \[ y = \cosh^{-1}\left(\frac{9x^{2}+10}{9}\right) = \cosh^{-1}\left(x^2 + \frac{10}{9}\right) \]
Step 2: Differentiating with respect to \( x \).
The derivative of \( \cosh^{-1}(u) \) is \( \frac{1}{\sqrt{u^2-1}} \cdot \frac{du}{dx} \). Let \( u = x^2 + \frac{10}{9} \): \[ \frac{du}{dx} = 2x \] Substituting into the formula: \[ \frac{dy}{dx} = \frac{1}{\sqrt{\left(x^2 + \frac{10}{9}\right)^2 - 1}} \cdot 2x \]
Step 3: Simplifying the algebraic denominator expression.
Expand the expression inside the square root: \[ \left(x^2 + \frac{10}{9}\right)^2 - 1 = \left(x^2 + \frac{10}{9} - 1\right)\left(x^2 + \frac{10}{9} + 1\right) = \left(x^2 + \frac{1}{9}\right)\left(x^2 + \frac{19}{9}\right) \] \[ = \frac{(9x^2+1)(9x^2+19)}{81} \] Taking the square root brings a factor of 9 to the numerator: \[ \frac{dy}{dx} = \frac{2x}{\frac{\sqrt{(9x^2+1)(9x^2+19)}}{9}} = \frac{18x}{\sqrt{(9x^2+19)(9x^2+1)}} \]
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