Exercise$\frac{x^2+x-2}{x^2+5x+6}$
Step-by-step Solution
1
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
$\frac{\frac{d}{dx}\left(x^2+x-2\right)\left(x^2+5x+6\right)-\left(x^2+x-2\right)\frac{d}{dx}\left(x^2+5x+6\right)}{\left(x^2+5x+6\right)^2}$
2
Simplify the product $-(x^2+x-2)$
$\frac{\frac{d}{dx}\left(x^2+x-2\right)\left(x^2+5x+6\right)+\left(-x^2-\left(x-2\right)\right)\frac{d}{dx}\left(x^2+5x+6\right)}{\left(x^2+5x+6\right)^2}$
3
Simplify the product $-(x-2)$
$\frac{\frac{d}{dx}\left(x^2+x-2\right)\left(x^2+5x+6\right)+\left(-x^2-x+2\right)\frac{d}{dx}\left(x^2+5x+6\right)}{\left(x^2+5x+6\right)^2}$
Intermediate steps
4
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{\left(\frac{d}{dx}\left(x^2\right)+1\right)\left(x^2+5x+6\right)+\left(-x^2-x+2\right)\frac{d}{dx}\left(x^2+5x+6\right)}{\left(x^2+5x+6\right)^2}$
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Intermediate steps
5
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{\left(\frac{d}{dx}\left(x^2\right)+1\right)\left(x^2+5x+6\right)+\left(-x^2-x+2\right)\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(5x\right)\right)}{\left(x^2+5x+6\right)^2}$
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Intermediate steps
6
The derivative of the linear function times a constant, is equal to the constant
$\frac{\left(\frac{d}{dx}\left(x^2\right)+1\right)\left(x^2+5x+6\right)+\left(-x^2-x+2\right)\left(\frac{d}{dx}\left(x^2\right)+5\frac{d}{dx}\left(x\right)\right)}{\left(x^2+5x+6\right)^2}$
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7
The derivative of the linear function is equal to $1$
$\frac{\left(\frac{d}{dx}\left(x^2\right)+1\right)\left(x^2+5x+6\right)+\left(-x^2-x+2\right)\left(\frac{d}{dx}\left(x^2\right)+5\right)}{\left(x^2+5x+6\right)^2}$
Intermediate steps
8
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{\left(2x+1\right)\left(x^2+5x+6\right)+\left(-x^2-x+2\right)\left(2x+5\right)}{\left(x^2+5x+6\right)^2}$
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Final answer to the exercise $\frac{\left(2x+1\right)\left(x^2+5x+6\right)+\left(-x^2-x+2\right)\left(2x+5\right)}{\left(x^2+5x+6\right)^2}$
