Question

find the derivative of log (x2+x+2/x2-x+2) w.r to x​

Answer 1

Step-by-step explanation:

ANSWER -2x+4/x^4+3x^2+4 (or) 4-2x/x^4+3x^2+4

Answer 2

Answer:

$\frac{-2x^{2}+2}{(x^{2}+x+1)(x^{2}-x+1)}$

Explanation:

Given: $\log_{e}(\frac{x^{2}+x+1}{x^{2}-x+1})$

Find: $\frac{d}{dx}\log_{e}(\frac{x^{2}+x+1}{x^{2}-x+1})$

Differentiation can be defined as a derivative of a function with respect to an independent variable.

Solution:

consider $\frac{d}{dx}\log_{e}(\frac{x^{2}+x+1}{x^{2}-x+1})$

Since, $\log_{e}(\frac{x^{2}+x+1}{x^{2}-x+1})=\ln (\frac{x^{2}+x+1}{x^{2}-x+1})$

$\frac{d}{dx}\log_{e}(\frac{x^{2}+x+1}{x^{2}-x+1})=\frac{d}{dx}\ln (\frac{x^{2}+x+1}{x^{2}-x+1})$

Apply Chain rule:

$\frac{d}{dx}\log_{e}(\frac{x^{2}+x+1}{x^{2}-x+1})=\frac{1}{\frac{x^{2}+x+1}{x^{2}-x+1}}\frac{d}{dx}\ln (\frac{x^{2}+x+1}{x^{2}-x+1})$

Since,$\frac{d}{dx}(\frac{x^{2}+x+1}{x^{2}-x+1})=\frac{-2x^{2}+2}{(x^{2}-x+1)^{2}}$

$\frac{d}{dx}\log_{e}(\frac{x^{2}+x+1}{x^{2}-x+1})=\frac{1}{\frac{x^{2}+x+1}{x^{2}-x+1}}\cdot \frac{-2x^{2}+2}{(x^{2}-x+1)^{2}}$

Hence, $\frac{d}{dx}\log_{e}(\frac{x^{2}+x+1}{x^{2}-x+1})=\frac{-2x^{2}+2}{(x^{2}+x+1)(x^{2}-x+1)}$

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