Answer to Question #134660 in Calculus for Tarun Joshua Daniel

Question

Differentiate the following with respect to x

e^x log x/ x+1

Answer 1

Not sure whether it is .../(x+1) or .../x + 1 so I'll just provide both.

1.

$$(\dfrac{e^x\log x}{x+1})' = \\ =\dfrac{(e^x\log x)'\cdot (x+1)-e^x\log x\cdot (x+1)'}{(x+1)^2}=\\ =\dfrac{(e^x\log x+\frac{e^x}{x})\cdot (x+1)-e^x\log x}{(x+1)^2}=\\ =\dfrac{xe^x\log x+e^x+\frac{e^x}{x}}{(x+1)^2}=e^x\cdot\dfrac{x\log x+1+\frac{1}{x}}{(x+1)^2}$$

2.

$$(\dfrac{e^x\log x}{x}+1)' = \\ =\dfrac{(e^x\log x)'\cdot x-e^x\log x\cdot (x)'}{x^2}=\\ =\dfrac{(e^x\log x+\frac{e^x}{x})\cdot x-e^x\log x}{x^2}=\\ =\dfrac{xe^x\log x+e^x-e^x\log x}{x^2}=\\ =e^x\cdot\dfrac{x\log x+1-\log x}{x^2}$$