Discussion Forum

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

Answer:

Explanation:

 U= $\cos h^{-1} x$ and v=log x

 $u=\log (x+\sqrt{x^{2}-1})$  and v=log x

 $\frac{du}{dx}=\left(\frac{1}{x+\sqrt{x^{2}-1}}\right)\left(1+\frac{x}{\sqrt{x^{2}-1}}\right)$  and

$\frac{dv}{dx}=\frac{1}{x}$

$\frac{du}{dx}$= $\frac{1}{\sqrt{x^{2}-1}}$  and $\frac{dv}{dx}=\frac{1}{x} \Rightarrow  \frac{du}{dv}=$

$\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{x}{\sqrt{x^{2}-1}}$

 $\because\left(\frac{du}{dv}\right)_{x=5}=\frac{5}{\sqrt{25-1}}=\frac{5}{\sqrt{24}}=\frac{5}{2\sqrt{6}}$