Q 24. - Calculus

Question

Find the derivatives of the functions: $f\left(x\right) = \frac{e^x \ln x}{x^2 - 1}$

Step-by-Step Solution

Answer

The required answer is $\frac{\left(x^2-1\right)\left(e^x{\displaystyle \frac{1}{x}}+e^x\ln x\right)+2x{e}^x\ln x}{{\left(x^2-1\right)}^2}$.

Step 1. Given information.

The given function is: $f\left(x\right) = \frac{e^x \ln x}{x^2 - 1}$

Step 2. Find the derivative of the given function.

$$\begin{gathered} f\left(x\right) = \frac{e^x \ln x}{x^2 - 1} \\ f\text{'}\left(x\right) =\frac{d}{dx} \left(\frac{e^x \ln x}{x^2 - 1}\right) \\ =\frac{\left(x^2 - 1\right){\displaystyle \frac{d}{dx}}\left(e^x\ln x\right)+e^x\ln x{\displaystyle \frac{d}{dx}}\left(x^2 - 1\right)}{{\left(x^2 - 1\right)}^2} \\ =\frac{\left(x^2 - 1\right)\left[e^x\frac{d}{dx} \ln x+\ln x \frac{d}{dx}{e}^x\right]+e^x\ln x 2x }{{\left(x^2 - 1\right)}^2} \\ =\frac{\left(x^2 - 1\right)\left[e^x{\displaystyle \frac{1}{x}}+e^{x }\ln x \right]+2x{e}^x\ln x }{{\left(x^2 - 1\right)}^2} \end{gathered}$$