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Here, we show you a step-by-step solved example of basic differentiation rules. 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
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{2-1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
Add the values $2$ and $-1$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{2-1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
Subtract the values $2$ and $-1$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
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The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)^{1}\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
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3
Any expression to the power of $1$ is equal to that same expression
$2\left(\frac{x^2+3x+1}{x^2+2x+2}\right)\frac{d}{dx}\left(\frac{x^2+3x+1}{x^2+2x+2}\right)$
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Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
$\frac{x^2+3x+1}{x^2+2x+2}\frac{2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^2}$
Intermediate steps
Multiplying fractions $\frac{x^2+3x+1}{x^2+2x+2} \times \frac{2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^2}$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)\left(x^2+2x+2\right)^2}$
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Multiplying fractions $\frac{x^2+3x+1}{x^2+2x+2} \times \frac{2\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^2}$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)\left(x^2+2x+2\right)^2}$
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Intermediate steps
When multiplying exponents with same base you can add the exponents: $\left(x^2+2x+2\right)\left(x^2+2x+2\right)^2$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{2+1}}$
Add the values $2$ and $1$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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When multiplying exponents with same base you can add the exponents: $\left(x^2+2x+2\right)\left(x^2+2x+2\right)^2$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)-\left(x^2+3x+1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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7
Simplify the product $-(x^2+3x+1)$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)+\left(-x^2-\left(3x+1\right)\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Simplify the product $-(3x+1)$
$\frac{2\left(x^2+3x+1\right)\left(\frac{d}{dx}\left(x^2+3x+1\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
Intermediate steps
The derivative of the constant function ($1$) is equal to zero
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\frac{d}{dx}\left(x^2+2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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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
The derivative of the constant function ($2$) is equal to zero
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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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
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$3\frac{d}{dx}\left(x\right)$
The derivative of the linear function is equal to $1$
$3$
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The derivative of the linear function times a constant, is equal to the constant
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+3\frac{d}{dx}\left(x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Intermediate steps
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$2\frac{d}{dx}\left(x\right)$
The derivative of the linear function is equal to $1$
$2$
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The derivative of the linear function times a constant, is equal to the constant
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+3\frac{d}{dx}\left(x\right)\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(x\right)\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Intermediate steps
The derivative of the linear function is equal to $1$
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+3\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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The derivative of the linear function is equal to $1$
$\frac{2\left(x^2+3x+1\right)\left(\left(\frac{d}{dx}\left(x^2\right)+3\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(\frac{d}{dx}\left(x^2\right)+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Intermediate steps
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2x^{\left(2-1\right)}$
Subtract the values $2$ and $-1$
$2x$
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The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{2\left(x^2+3x+1\right)\left(\left(2x+3\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$
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Final answer to the problem
$\frac{2\left(x^2+3x+1\right)\left(\left(2x+3\right)\left(x^2+2x+2\right)+\left(-x^2-3x-1\right)\left(2x+2\right)\right)}{\left(x^2+2x+2\right)^{3}}$