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Wize High School Grade 12 Calculus Textbook > Derivatives of Polynomials

Quotient Rule

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Quotient Rule

If h(x)=f(x)g(x)\displaystyle h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)} and g(x)≠0g\left(x\right)\ne0, then h′(x)=f′(x)g(x)−f(x)g′(x)[g(x)]2\displaystyle h'\left(x\right)=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{\left[g\left(x\right)\right]^2}

Example
Find the derivative of f(x)=x3+72x+5\displaystyle f\left(x\right)=\frac{x^3+7}{2x+5}.

f′(x)=(x3+7)′(2x+5)−(x3+7)(2x+5)′[(2x+5)]2\displaystyle f'\left(x\right)=\frac{\orange{\left(x^3+7\right)'}\left(2x+5\right)-\left(x^3+7\right)\green{\left(2x+5\right)'}}{\left[\left(2x+5\right)\right]^2}
f′(x)=(3x2)(2x+5)−(x3+7)(2)[(2x+5)]2\displaystyle f'\left(x\right)=\frac{\orange{\left(3x^2\right)}\left(2x+5\right)-\left(x^3+7\right)\green{\left(2\right)}}{\left[\left(2x+5\right)\right]^2}

Simplifying:
f′(x)=3x2(2x+5)−2(x3+7)(2x+5)2\displaystyle f'\left(x\right)=\frac{3x^2\left(2x+5\right)-2\left(x^3+7\right)}{\left(2x+5\right)^2}
f′(x)=6x3+15x2−2x3−14(2x+5)2\displaystyle f'\left(x\right)=\frac{6x^3+15x^2-2x^3-14}{\left(2x+5\right)^2}
f′(x)=4x3+15x2−14(2x+5)2\displaystyle f'\left(x\right)=\frac{4x^3+15x^2-14}{\left(2x+5\right)^2}
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Example: Quotient Rule

Find the derivative of y=2x4−7x+1x3y=\frac{2x^4-7x+1}{x^3} using 3 different methods.

Method 1: Using Quotient Rule
y′ or dydx=(2x4−7x+1)′(x3)−(2x4−7x+1)(x3)′(x3)2y'\ or\ \frac{dy}{dx}=\frac{\orange{\left(2x^4-7x+1\right)'}\left(x^3\right)-\left(2x^4-7x+1\right)\green{\left(x^3\right)'}}{\left(x^3\right)^2}
y′ or dydx=(8x3−7)′(x3)−(2x4−7x+1)(3x2)′(x3)2y'\ or\ \frac{dy}{dx}=\frac{\orange{\left(8x^3-7\right)'}\left(x^3\right)-\left(2x^4-7x+1\right)\green{\left(3x^2\right)'}}{\left(x^3\right)^2}
Simplify:
y′ or dydx=x3(8x3−7)−3x2(2x4−7x+1)x6y'\ or\ \frac{dy}{dx}=\frac{x^3\left(8x^3-7\right)-3x^2\left(2x^4-7x+1\right)}{x^6}
y′ or dydx=8x6−7x3−6x6+21x3−3x2x6y'\ or\ \frac{dy}{dx}=\frac{8x^6-7x^3-6x^6+21x^3-3x^2}{x^6}
y′ or dydx=2x6+14x3−3x2x6y'\ or\ \frac{dy}{dx}=\frac{2x^6+14x^3-3x^2}{x^6}
y′ or dydx=2x4+14x−3x4y'\ or\ \frac{dy}{dx}=\frac{2x^4+14x-3}{x^4}

Method 2: Using Product Rule
y=(2x4−7x+1)(x−3)y=\left(2x^4-7x+1\right)\left(x^{-3}\right)
y′ or dydx=(2x4−7x+1)′(x−3)+(2x4−7x+1)(x−3)′y'\ or\ \frac{dy}{dx}=\orange{\left(2x^4-7x+1\right)'}\left(x^{-3}\right)+\left(2x^4-7x+1\right)\green{\left(x^{-3}\right)'}
y′ or dydx=(8x3−7)(x−3)+(2x4−7x+1)(−3x−4)y'\ or\ \frac{dy}{dx}=\orange{\left(8x^3-7\right)}\left(x^{-3}\right)+\left(2x^4-7x+1\right)\green{\left(-3x^{-4}\right)}
Simplify:
y′ or dydx=8x3−7x3+−3(2x4−7x+1)x4y'\ or\ \frac{dy}{dx}=\frac{8x^3-7}{x^3}+\frac{-3\left(2x^4-7x+1\right)}{x^4}
y′ or dydx=8x3−7x3+−6x4+21x+−3x4y'\ or\ \frac{dy}{dx}=\frac{8x^3-7}{x^3}+\frac{-6x^4+21x+-3}{x^4}
y′ or dydx=8x4−7xx4+−6x4+21x+−3x4y'\ or\ \frac{dy}{dx}=\frac{8x^4-7x}{x^4}+\frac{-6x^4+21x+-3}{x^4}
y′ or dydx=2x4+14x−3x4y'\ or\ \frac{dy}{dx}=\frac{2x^4+14x-3}{x^4}

Method 3: Simplify then use Power Rule
y=2x4x3−7xx3+1x3y=\frac{2x^4}{x^3}-\frac{7x}{x^3}+\frac{1}{x^3}
y=2x−7x2+1x3y=2x-\frac{7}{x^2}+\frac{1}{x^3}
y=2x−7x−2+x−3y=2x-7x^{-2}+x^{-3}
y′ or dydx=2+14x−3−3x−4y'\ or\ \frac{dy}{dx}=2+14x^{-3}-3x^{-4}
Simplify:
y′ or dydx=2+14x3−3x4y'\ or\ \frac{dy}{dx}=2+\frac{14}{x^3}-\frac{3}{x^4}
y′ or dydx=2x4+14x−3x4y'\ or\ \frac{dy}{dx}=\frac{2x^4+14x-3}{x^4}

Find the derivative of the following functions at the given point using quotient rule.


*You may also need to use other derivative rules you've learned so far.
Derivative of y=2x5−4x(4−x2)3\displaystyle y=\frac{2x^5-4x}{\left(4-x^2\right)^3}, at x=0x=0.

Practice: Quotient Rule Application Problem

The graph of y=ax−b(x−2)(x−3)\displaystyle y=\frac{ax-b}{\left(x-2\right)\left(x-3\right)} has a horizontal tangent line at the point (1, −13)\left(1,\ -\frac{1}{3}\right). Find the values of aa and bb.

Example: Product & Quotient Rule

Suppose g(x)g\left(x\right) is differentiable, f(x)=1+x g(x)x2f\left(x\right)=\frac{1+x\ g\left(x\right)}{x^2}, f′(1)=2f'\left(1\right)=2, g(1)=1g\left(1\right)=1, what is g′(1)g'\left(1\right)?
☐ 0
☐ 1
☐ 5
☐ -5
☐ None of the above
We can find the derivative of ff using product and quotient rules.
f′(x)=[1+x g(x)]′[x2]−[1+x g(x)][x2]′[x2]2\displaystyle f'\left(x\right)= \frac{\textcolor{orange}{\left[1+x\ g\left(x\right)\right]'} \textcolor{blue}{\left[x^2\right]}- \green{\left[1+x\ g\left(x\right)\right]}\purple{\left[x^2\right]'}}{\left[x^2\right]^2}

f′(x)=[0+x′g(x)+xg′(x)][x2]−[1+x g(x)][2x][x2]2\displaystyle f'\left(x\right)= \frac{\textcolor{orange}{\left[0+x' g(x)+x g'(x)\right]} \textcolor{blue}{\left[x^2\right]}- \green{\left[1+x\ g\left(x\right)\right]}\purple{\left[2x\right]}}{\left[x^2\right]^2}
f′(x)=[0+g(x)+xg′(x)][x2]−[1+x g(x)][2x][x2]2\displaystyle f'\left(x\right)= \frac{\textcolor{orange}{\left[0+g(x)+x g'(x)\right]} \textcolor{blue}{\left[x^2\right]}- \green{\left[1+x\ g\left(x\right)\right]}\purple{\left[2x\right]}}{\left[x^2\right]^2}

f′(x)=[g(x)+x g′(x)][x2]−[1+x g(x)][2x]x4\displaystyle f'\left(x\right)=\frac{\left[g\left(x\right)+x\ g'\left(x\right)\right]\left[x^2\right]-\left[1+x\ g\left(x\right)\right]\left[2x\right]}{x^4}

Sub in x=1x=1:
f′(1)=[g(1)+1 g′(1)][12]−[1+1 g(1)][2⋅1]14\displaystyle f'\left(1\right)=\frac{\left[g\left(1\right)+1\ g'\left(1\right)\right]\left[1^2\right]-\left[1+1\ g\left(1\right)\right]\left[2\cdot1\right]}{1^4}

f′(1)=g(1)+g′(1)−[1+g(1)][2]f'\left(1\right)=g\left(1\right)+g'\left(1\right)-\left[1+g\left(1\right)\right]\left[2\right]

f′(1)=g(1)+g′(1)−2−2g(1)f'\left(1\right)=g\left(1\right)+g'\left(1\right)-2-2g\left(1\right)

Sub in the known values:
2=1+g′(1)−2−2(1)2=1+g'\left(1\right)-2-2\left(1\right)
g′(1)=5g'\left(1\right)=5

Alternative method:
We can first simplify the expression:
f(x)=1x2+x g(x)x2\displaystyle f\left(x\right)=\frac{1}{x^2}+\frac{x\ g\left(x\right)}{x^2}

f(x)=x−2+g(x)x\displaystyle f\left(x\right)=x^{-2}+\frac{g\left(x\right)}{x}

Now we can find the derivative with just our regular derivative rules and the quotient rule
f′(x)=−2x−3+(g′(x))(x)−(1)(g(x))x2\displaystyle f'\left(x\right)=-2x^{-3}+\frac{\left(g'\left(x\right)\right)\left(x\right)-\left(1\right)\left(g\left(x\right)\right)}{x^2}

f′(x)=−2x3+g′(x)(x)−g(x)x2\displaystyle f'\left(x\right)=-\frac{2}{x^3}+\frac{g'\left(x\right)\left(x\right)-g\left(x\right)}{x^2}

Substitute x=1:
f′(1)=−213+g′(1)(1)−g(1)12\displaystyle f'\left(1\right)=-\frac{2}{1^3}+\frac{g'\left(1\right)\left(1\right)-g\left(1\right)}{1^2}

f′(1)=−2+g′(1)−g(1)f'\left(1\right)=-2+g'\left(1\right)-g\left(1\right)

Substitute known values into this expression:
2=−2+g′(1)−12=-2+g'\left(1\right)-1
2=−3+g′(1)2=-3+g'\left(1\right)
5=g′(1)5=g'\left(1\right)

So, g′(1)=5g'\left(1\right)=5