1
Here, we show you a step-by-step solved example of règles de différenciation de base. This solution was automatically generated by our smart calculator:
$\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^2$
Intermediate steps
Apply the formula: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, where $a=2$ and $x=\frac{x^2+3x+1}{x^2+2x+2}$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{2-1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
Apply the formula: $a+b$$=a+b$, where $a=2$, $b=-1$ and $a+b=2-1$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
Apply the formula: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, where $a=2$ and $x=\frac{x^2+3x+1}{x^2+2x+2}$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{2-1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
Apply the formula: $a+b$$=a+b$, where $a=2$, $b=-1$ and $a+b=2-1$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
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Apply the formula: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, where $a=2$ and $x=\frac{x^2+3x+1}{x^2+2x+2}$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
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3
Apply the formula: $x^1$$=x$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
4
Apply the formula: $\frac{d}{dx}\left(\frac{a}{b}\right)$$=\frac{\frac{d}{dx}\left(a\right)b-a\frac{d}{dx}\left(b\right)}{b^2}$, where $a=x^2+3x+1$ and $b=x^2+2x+2$
$\frac{x^2+3x+1}{x^2+2x+2}\frac{2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^2}$
Intermediate steps
Apply the formula: $\frac{a}{b}\frac{c}{f}$$=\frac{ac}{bf}$, where $a=x^2+3x+1$, $b=x^2+2x+2$, $c=2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)$, $a/b=\frac{x^2+3x+1}{x^2+2x+2}$, $f=\left(x^2+2x+2\right)^2$, $c/f=\frac{2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^2}$ and $a/bc/f=\frac{x^2+3x+1}{x^2+2x+2}\frac{2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^2}$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)\left(x^2+2x+2\right)^2}$
5
Apply the formula: $\frac{a}{b}\frac{c}{f}$$=\frac{ac}{bf}$, where $a=x^2+3x+1$, $b=x^2+2x+2$, $c=2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)$, $a/b=\frac{x^2+3x+1}{x^2+2x+2}$, $f=\left(x^2+2x+2\right)^2$, $c/f=\frac{2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^2}$ and $a/bc/f=\frac{x^2+3x+1}{x^2+2x+2}\frac{2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^2}$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)\left(x^2+2x+2\right)^2}$
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Intermediate steps
Apply the formula: $x\cdot x^n$$=x^{\left(n+1\right)}$, where $x^nx=\left(x^2+2x+2\right)\left(x^2+2x+2\right)^2$, $x=x^2+2x+2$, $x^n=\left(x^2+2x+2\right)^2$ and $n=2$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{2+1}}$
Apply the formula: $a+b$$=a+b$, where $a=2$, $b=1$ and $a+b=2+1$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
6
Apply the formula: $x\cdot x^n$$=x^{\left(n+1\right)}$, where $x^nx=\left(x^2+2x+2\right)\left(x^2+2x+2\right)^2$, $x=x^2+2x+2$, $x^n=\left(x^2+2x+2\right)^2$ and $n=2$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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7
Apply the formula: $-\left(a+b\right)$$=-a-b$, where $a=x^2$, $b=3x+1$, $-1.0=-1$ and $a+b=x^2+3x+1$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)+\left(-x^2-\left(3x+1\right)\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
8
Apply the formula: $-\left(a+b\right)$$=-a-b$, where $a=3x$, $b=1$, $-1.0=-1$ and $a+b=3x+1$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
Intermediate steps
Apply the formula: $\frac{d}{dx}\left(c\right)$$=0$, where $c=1$
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
9
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Intermediate steps
Apply the formula: $\frac{d}{dx}\left(c\right)$$=0$, where $c=2$
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)\right)}{\left(x^2+2x+2\right)^{3}}$
10
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Intermediate steps
Apply the formula: $\frac{d}{dx}\left(cx\right)$$=c\frac{d}{dx}\left(x\right)$
$3\frac{d}{dx}\left(x\right)$
Apply the formula: $\frac{d}{dx}\left(x\right)$$=1$
$3$
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Apply the formula: $\frac{d}{dx}\left(nx\right)$$=n\frac{d}{dx}\left(x\right)$, where $n=3$
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+3\frac{d}{dx}\left(x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Intermediate steps
Apply the formula: $\frac{d}{dx}\left(cx\right)$$=c\frac{d}{dx}\left(x\right)$
$2\frac{d}{dx}\left(x\right)$
Apply the formula: $\frac{d}{dx}\left(x\right)$$=1$
$2$
12
Apply the formula: $\frac{d}{dx}\left(nx\right)$$=n\frac{d}{dx}\left(x\right)$, where $n=2$
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+3\frac{d}{dx}\left(x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(x\right)\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Intermediate steps
Apply the formula: $\frac{d}{dx}\left(x\right)$$=1$
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+3\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
13
Apply the formula: $\frac{d}{dx}\left(x\right)$$=1$
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+3\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Intermediate steps
Apply the formula: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}$, where $a=2$
$2x^{\left(2-1\right)}$
Apply the formula: $a+b$$=a+b$, where $a=2$, $b=-1$ and $a+b=2-1$
$2x$
14
Apply the formula: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}$, where $a=2$
$\frac{2\left(x^2+3x+1\right)\left(\left(2x+3\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Final answer to the problem
$\frac{2\left(x^2+3x+1\right)\left(\left(2x+3\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$