Question

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

Short 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-by-step solution

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.

$$\begin{gathered} f\left(x\right)=\frac{e^x\ln x}{x^2-1} \\ f\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{e^x\ln x}{x^2-1}\right) \\ =\frac{\left(x^2-1\right){\displaystyle \frac{d}{dx}}\left(e^x\ln x\right)+e^x\ln x{\displaystyle \frac{d}{dx}}\left(x^2-1\right)}{{\left(x^2-1\right)}^2} \\ =\frac{\left(x^2-1\right)\left[e^x\frac{d}{dx}\ln x+\ln x\frac{d}{dx}{e}^x\right]+e^x\ln x2x}{{\left(x^2-1\right)}^2} \\ =\frac{\left(x^2-1\right)\left[e^x{\displaystyle \frac{1}{x}}+e^x\ln x\right]+2x{e}^x\ln x}{{\left(x^2-1\right)}^2} \end{gathered}$$