(A) Let $u = \cosh^{-1} x$. The derivative is $\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$.Let $v = \log x$. The derivative is $\frac{dv}{dx} = \frac{1}{x

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(A) Let $u = \cosh^{-1} x$. The derivative is $\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$.Let $v = \log x$. The derivative is $\frac{dv}{dx} = \frac{1}{x}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/\sqrt{x^2-1}}{1/x} = \frac{x}{\sqrt{x^2-1}}$.At $x=5$,$\frac{du}{dv} = \frac{5}{\sqrt{5^2-1}} = \frac{5}{\sqrt{25-1}} = \frac{5}{\sqrt{24}}$.

Class 12 Mathematics Continuity and Differentiation Derivative at a point, Standard differentiation

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The derivative of $\cosh^{-1} x$ with respect to $\log x$ at $x=5$ is

TS EAMCET 2019 All TS EAMCET

A
$\frac{5}{\sqrt{24}}$
B
$\frac{1}{\sqrt{24}}$
C
$\frac{5}{\sqrt{26}}$
D
$\frac{1}{\sqrt{26}}$

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Derivative at a point, Standard differentiation

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