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\frac{d}{dx}\left(\left(\frac{3x-1}{x^2+3}\right)^2\right)

Derive the function ((3x-1)/(x^2+3))^2 with respect to x

Answer

$\frac{6x\frac{3\left(3+x^2\right)-2x\left(3x-1\right)}{\left(3+x^2\right)^2}-2\frac{3\left(3+x^2\right)-2x\left(3x-1\right)}{\left(3+x^2\right)^2}}{3+x^2}$

Step-by-step explanation

Problem

$\frac{d}{dx}\left(\left(\frac{3x-1}{x^2+3}\right)^2\right)$
1

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$2\frac{3x-1}{3+x^2}\cdot\frac{d}{dx}\left(\frac{3x-1}{3+x^2}\right)$

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Answer

$\frac{6x\frac{3\left(3+x^2\right)-2x\left(3x-1\right)}{\left(3+x^2\right)^2}-2\frac{3\left(3+x^2\right)-2x\left(3x-1\right)}{\left(3+x^2\right)^2}}{3+x^2}$

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$\frac{d}{dx}\left(\left(\frac{3x-1}{x^2+3}\right)^2\right)$

Main topic:

Differential calculus

Used formulas:

4. See formulas

Time to solve it:

~ 0.78 seconds