Wize High School Grade 12 Calculus Textbook > Derivatives of Polynomials

Product Rule

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Product Rule
If h(x)=f(x)g(x)h\left(x\right)=f\left(x\right)g\left(x\right)h(x)=f(x)g(x), then h′(x)=f′(x) g(x)+f(x) g′(x)h'\left(x\right)=f'\left(x\right)\ g\left(x\right)+f\left(x\right)\ g'\left(x\right)h′(x)=f′(x) g(x)+f(x) g′(x).

Example
Find the derivative of f(x)=(x+x2)(4x−1)f\left(x\right)=\left(\sqrt{x}+x^2\right)\left(4x-1\right)f(x)=(x​+x2)(4x−1) using the product rule.

f(x)=(x12+x2)(4x−1)f\left(x\right)=\left(x^{\frac{1}{2}}+x^2\right)\left(4x-1\right)f(x)=(x21​+x2)(4x−1)
f′(x)=(x12+x2)′ (4x−1)+(x12+x2)(4x−1)′f'\left(x\right)=\orange{\left(x^{\frac{1}{2}}+x^2\right)'}\ \left(4x-1\right)+\left(x^{\frac{1}{2}}+x^2\right)\green{\left(4x-1\right)'}f′(x)=(x21​+x2)′ (4x−1)+(x21​+x2)(4x−1)′
f′(x)=(12x−12+2x)(4x−1)+(x12+x2)(4)f'(x)=\orange{\left(\frac{1}{2}x^{-\frac{1}{2}}+2x\right)}\left(4x-1\right)+\left(x^{\frac{1}{2}}+x^2\right)\green{\left(4\right)}f′(x)=(21​x−21​+2x)(4x−1)+(x21​+x2)(4)
f′(x)=(12x+2x)(4x−1)+(x+x2)(4)f'(x)=\left(\frac{1}{2\sqrt{x}}+2x\right)\left(4x-1\right)+\left(\sqrt{x}+x^2\right)\left(4\right)f′(x)=(2x​1​+2x)(4x−1)+(x​+x2)(4)

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Power of a Function Rule
If h(x)=[f(x)]nh\left(x\right)=\left[f\left(x\right)\right]^nh(x)=[f(x)]n, then h′(x)=n[f(x)]n−1×f′(x)h'\left(x\right)=n\left[f\left(x\right)\right]^{n-1}\times f'\left(x\right)h′(x)=n[f(x)]n−1×f′(x).

Example
Find the derivative of f(x)=(3x2−1)5f\left(x\right)=\left(3x^2-1\right)^5f(x)=(3x2−1)5 using the generalized power rule.

f′(x)=5[(3x2−1)]4×(3x2−1)′f'\left(x\right)=5\left[\left(3x^2-1\right)\right]^4\times\orange{\left(3x^2-1\right)'}f′(x)=5[(3x2−1)]4×(3x2−1)′
f′(x)=5(3x2−1)4×(6x)f'(x)=5\left(3x^2-1\right)^4\times\orange{\left(6x\right)}f′(x)=5(3x2−1)4×(6x)
f′(x)=30x(3x2−1)4f'(x)=30x\left(3x^2-1\right)^4f′(x)=30x(3x2−1)4

Practice: Product Rule
Use the product rule to differentiate each of the following functions.
a) f(x)=(x2−1)(2−x3)f\left(x\right)=\left(x^2-1\right)\left(2-x^3\right)f(x)=(x2−1)(2−x3)
b) g(x)=(x+1)(3x2−x+2)g\left(x\right)=\left(x+1\right)\left(3x^2-x+2\right)g(x)=(x+1)(3x2−x+2)

f'(x)=

g'(x)=

I don't know

Practice: Power of a Function Rule
Find the derivative of f(x)=(1x+2x3)4f\left(x\right)=\left(\frac{1}{x}+2x^3\right)^4f(x)=(x1​+2x3)4 at the point x=1x=1x=1

Enter the exact value.

f'(1)=

I don't know

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Example: Product & Power of a Function Rule
Determine the derivative of y=(3x2−4x+5)(x3−1)5y=\left(3x^2-4x+5\right)\left(x^3-1\right)^5y=(3x2−4x+5)(x3−1)5 at the point (1, 0)\left(1,\ 0\right)(1, 0).

Since we see two terms involving xxxs being multiplied together, we need to use the product rule to find the derivative:
y′=(3x2−4x+5)′(x3−1)5+(3x2−4x+5)((x3−1)5)′y'=\orange{\left(3x^2-4x+5\right)'}\left(x^3-1\right)^5+\left(3x^2-4x+5\right)\green{\left(\left(x^3-1\right)^5\right)'}y′=(3x2−4x+5)′(x3−1)5+(3x2−4x+5)((x3−1)5)′
y′=(6x−4)(x3−1)5+(3x2−4x+5)[5(x3−1)4×3x2]y'=\orange{\left(6x-4\right)}\left(x^3-1\right)^5+\left(3x^2-4x+5\right)\green{\left[5\left(x^3-1\right)^4\times3x^2\right]}y′=(6x−4)(x3−1)5+(3x2−4x+5)[5(x3−1)4×3x2]

We can simplify this a bit:
y′=(6x−4)(x3−1)5+15x2(3x2−4x+5)(x3−1)4y'=\left(6x-4\right)\left(x^3-1\right)^5+15x^2\left(3x^2-4x+5\right)\left(x^3-1\right)^4y′=(6x−4)(x3−1)5+15x2(3x2−4x+5)(x3−1)4

Sub in the value x=1x=1x=1:
y′(1)=(6(1)−4)((1)3−1)5+15(1)2(3(1)2−4(1)+5)((1)3−1)4y'\left(1\right)=\left(6\left(1\right)-4\right)\left(\left(1\right)^3-1\right)^5+15\left(1\right)^2\left(3\left(1\right)^2-4\left(1\right)+5\right)\left(\left(1\right)^3-1\right)^4y′(1)=(6(1)−4)((1)3−1)5+15(1)2(3(1)2−4(1)+5)((1)3−1)4
y′(1)=0+0y'\left(1\right)=0+0y′(1)=0+0
y′(0)=0y'\left(0\right)=0y′(0)=0

Practice: Product & Power of a Function Rule
Find the rate of change of f(x)=(3−2x5)3(4x−1)f\left(x\right)=\left(3-2x^5\right)^3\left(4x-1\right)f(x)=(3−2x5)3(4x−1) at the point (0, −3)\left(0,\ -3\right)(0, −3).

Enter the exact value of the rate of change

I don't know

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Example: Product & Power of a Function Rule Application
Find the equation of the tangent line to the curve y=(3x4−2x2+1)2(x3−2)5y=\left(3x^4-2x^2+1\right)^2\left(x^3-2\right)^5y=(3x4−2x2+1)2(x3−2)5 at the point x=1x=1x=1.

When x=1x=1x=1, y=(3(1)4−2(1)2+1)2((1)3−2)5=−4y=\left(3\left(1\right)^4-2\left(1\right)^2+1\right)^2\left(\left(1\right)^3-2\right)^5=-4y=(3(1)4−2(1)2+1)2((1)3−2)5=−4. So a point on the tangent line is (1, −4)\left(1,\ -4\right)(1, −4).

The slope of this tangent line is the derivative of the curve at the point x=1x=1x=1:
y′=((3x4−2x2+1)2)′(x3−2)5+(3x4−2x2+1)2((x3−2)5)′y'=\orange{\left(\left(3x^4-2x^2+1\right)^2\right)'}\left(x^3-2\right)^5+\left(3x^4-2x^2+1\right)^2\green{\left(\left(x^3-2\right)^5\right)'}y′=((3x4−2x2+1)2)′(x3−2)5+(3x4−2x2+1)2((x3−2)5)′
y′=(2(3x4−2x2+1)(12x3−4x))(x3−2)5+(3x4−2x2+1)2[5(x3−2)4(3x2)]y'=\orange{\left(2\left(3x^4-2x^2+1\right)\left(12x^3-4x\right)\right)}\left(x^3-2\right)^5+\left(3x^4-2x^2+1\right)^2\green{\left[5\left(x^3-2\right)^4\left(3x^2\right)\right]}y′=(2(3x4−2x2+1)(12x3−4x))(x3−2)5+(3x4−2x2+1)2[5(x3−2)4(3x2)]
y′=(24x3−8x)(3x4−2x2+1)(x3−2)5+15x2(3x4−2x2+1)2(x3−2)4y'=\left(24x^3-8x\right)\left(3x^4-2x^2+1\right)\left(x^3-2\right)^5+15x^2\left(3x^4-2x^2+1\right)^2\left(x^3-2\right)^4y′=(24x3−8x)(3x4−2x2+1)(x3−2)5+15x2(3x4−2x2+1)2(x3−2)4
y′(1)=(24(1)3−8(1))(3(1)4−2(1)2+1)((1)3−2)5+15(1)2(3(1)4−2(1)2+1)2((1)3−2)4y'\left(1\right)=\left(24\left(1\right)^3-8\left(1\right)\right)\left(3\left(1\right)^4-2\left(1\right)^2+1\right)
\left(\left(1\right)^3-2\right)^5+15\left(1\right)^2\left(3\left(1\right)^4-2\left(1\right)^2+1\right)^2\left(\left(1\right)^3-2\right)^4y′(1)=(24(1)3−8(1))(3(1)4−2(1)2+1)((1)3−2)5+15(1)2(3(1)4−2(1)2+1)2((1)3−2)4
y′(1)=(16)(2)(−1)5+15(2)2(−1)4y'\left(1\right)=\left(16\right)\left(2\right)\left(-1\right)^5+15\left(2\right)^2\left(-1\right)^4y′(1)=(16)(2)(−1)5+15(2)2(−1)4
y′(1)=−32+60y'\left(1\right)=-32+60y′(1)=−32+60
y′(1)=28y'\left(1\right)=28y′(1)=28

Therefore, the equation of the tangent line is:
y−y0=m(x−x0)y-y_0=m\left(x-x_0\right)y−y0​=m(x−x0​)
y−(−4)=28(x−1)y-\left(-4\right)=28\left(x-1\right)y−(−4)=28(x−1)
y+4=28x−28y+4=28x-28y+4=28x−28
y=28x−32y=28x-32y=28x−32

Practice: Product & Power of a Function Rule Application
How many horizontal tangent lines are there to the curve y=(3x+1)3(x−2)3y=\left(3x+1\right)^3\left(x-2\right)^3y=(3x+1)3(x−2)3 ?

The curve does not have any horizontal tangent lines

The curve has 1 horizontal tangent line only

The curve has 2 horizontal tangent lines

The curve has 3 horizontal tangent lines

The curve has 4 horizontal tangent lines

I don't know

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Example: Product Rule & Rational Functions
Find the derivative of y=x+13x−2\displaystyle y=\frac{x+1}{3x-2}y=3x−2x+1​ by first rewriting it as a product of two functions.

Rewrite
y=(x+1)(13x−2)y=\left(x+1\right)\left(\frac{1}{3x-2}\right)y=(x+1)(3x−21​)
y=(x+1)(3x−2)−1y=\left(x+1\right)\left(3x-2\right)^{-1}y=(x+1)(3x−2)−1

Using the product rule:
y′=(x+1)′(3x−2)−1+(x+1)[(3x−2)−1]′y'=\orange{\left(x+1\right)'}\left(3x-2\right)^{-1}+\left(x+1\right)\green{\left[\left(3x-2\right)^{-1}\right]'}y′=(x+1)′(3x−2)−1+(x+1)[(3x−2)−1]′
*We need to use the power of a function derivative rule for the derivative of the second term
y′=(1)(3x−2)−1+(x+1)[−1(3x−2)−2×(3x−2)′]y'=\orange{\left(1\right)}\left(3x-2\right)^{-1}+\left(x+1\right)\green{\left[-1\left(3x-2\right)^{-2}\times\pink{\left(3x-2\right)'}\right]}y′=(1)(3x−2)−1+(x+1)[−1(3x−2)−2×(3x−2)′]
y′=(1)(3x−2)−1+(x+1)[−1(3x−2)−2×3]y'=\left(1\right)\left(3x-2\right)^{-1}+\left(x+1\right)\left[-1\left(3x-2\right)^{-2}\times\pink{3}\right]y′=(1)(3x−2)−1+(x+1)[−1(3x−2)−2×3]
y′=13x−2+(x+1)[−3(3x−2)2]\displaystyle y'=\frac{1}{3x-2}+\left(x+1\right)\left[-\frac{3}{\left(3x-2\right)^2}\right]y′=3x−21​+(x+1)[−(3x−2)23​]
y′=13x−2−3x+3(3x−2)2\displaystyle y'=\frac{1}{3x-2}-\frac{3x+3}{\left(3x-2\right)^2}y′=3x−21​−(3x−2)23x+3​
y′=3x−2(3x−2)2−3x+3(3x−2)2\displaystyle y'=\frac{3x-2}{\left(3x-2\right)^2}-\frac{3x+3}{\left(3x-2\right)^2}y′=(3x−2)23x−2​−(3x−2)23x+3​
y′=3x−2−3x−3(3x−2)2\displaystyle y'=\frac{3x-2-3x-3}{\left(3x-2\right)^2}y′=(3x−2)23x−2−3x−3​
y′=−5(3x−2)2\displaystyle y'=\frac{-5}{\left(3x-2\right)^2}y′=(3x−2)2−5​

Practice: Product Rule & Rational Functions
The portion of cells in a body that are affected by a certain virus is modelled by the function A(t)=t−1t+2\displaystyle A(t)=\frac{t-1}{t+2}A(t)=t+2t−1​ where ttt is the number of hours after the virus first enters the body, and  t≥1t\ge1t≥1.

Determine how quickly the percentage of cells in the body are being affected when 40% of the cells in this body are affected.

Enter the exact rate of change (portion/hour) of the portion of cells that are affected

I don't know

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Example: Product Rule for 3 Functions
Find the derivative of the following functions:
a) f(x)=(x−2)(3x2+x−2)(2x3−4)f\left(x\right)=\left(x-2\right)\left(3x^2+x-2\right)\left(2x^3-4\right)f(x)=(x−2)(3x2+x−2)(2x3−4)

Group two of these factors together:
f(x)=[(x−2)(3x2+x−2)][2x3−4]f\left(x\right)=\left[\left(x-2\right)\left(3x^2+x-2\right)\right]\left[2x^3-4\right]f(x)=[(x−2)(3x2+x−2)][2x3−4]

Now, we use the product rule to find the derivative:
f′(x)=[(x−2)(3x2+x−2)]′[2x3−4]+[(x−2)(3x2+x−2)][2x3−4]′f'\left(x\right)=\orange{\left[\left(x-2\right)\left(3x^2+x-2\right)\right]'}\left[2x^3-4\right]+\left[\left(x-2\right)\left(3x^2+x-2\right)\right]\green{\left[2x^3-4\right]'}f′(x)=[(x−2)(3x2+x−2)]′[2x3−4]+[(x−2)(3x2+x−2)][2x3−4]′

Use product rule again to find the derivative of the first part:
f′(x)=[(x−2)′(3x2+x−2)+(x−2)(3x2+x−2)′][2x3−4]+[(x−2)(3x2+x−2)][6x2]f'\left(x\right)=\orange{\left[\red{\left(x-2\right)'}\left(3x^2+x-2\right)+\left(x-2\right)\blue{\left(3x^2+x-2\right)'}\right]}\left[2x^3-4\right]+\left[\left(x-2\right)\left(3x^2+x-2\right)\right]\green{\left[6x^2\right]}f′(x)=[(x−2)′(3x2+x−2)+(x−2)(3x2+x−2)′][2x3−4]+[(x−2)(3x2+x−2)][6x2]
f′(x)=[(1)(3x2+x−2)+(x−2)(6x+1)][2x3−4]+[(x−2)(3x2+x−2)][6x2]f'\left(x\right)=\orange{\left[
\red{\left(1\right)}

\left(3x^2+x-2\right)+\left(x-2\right)
\blue{\left(6x+1\right)}\right]}
\left[2x^3-4\right]

+\left[\left(x-2\right)\left(3x^2+x-2\right)\right]
\green{\left[6x^2\right]}f′(x)=[(1)(3x2+x−2)+(x−2)(6x+1)][2x3−4]+[(x−2)(3x2+x−2)][6x2]

We can simplify this a bit:
f′(x)=[3x2+x−2+6x2+x−12x−2][2x3−4]+6x2(x−2)(3x2+x−2)f'\left(x\right)=\left[3x^2+x-2+6x^2+x-12x-2\right]\left[2x^3-4\right]+6x^2\left(x-2\right)\left(3x^2+x-2\right)f′(x)=[3x2+x−2+6x2+x−12x−2][2x3−4]+6x2(x−2)(3x2+x−2)
f′(x)=(9x2−10x−4)(2x3−4)+6x2(x−2)(3x2+x−2)f'\left(x\right)=\left(9x^2-10x-4\right)\left(2x^3-4\right)+6x^2\left(x-2\right)\left(3x^2+x-2\right)f′(x)=(9x2−10x−4)(2x3−4)+6x2(x−2)(3x2+x−2)

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b) g(x)=(x+2)(3−4x)3(2x2+1)2g\left(x\right)=\left(x+2\right)\left(3-4x\right)^3\left(2x^2+1\right)^2g(x)=(x+2)(3−4x)3(2x2+1)2

Group two of these factors together:
g(x)=[(x+2)(3−4x)3][(2x2+1)2]g\left(x\right)=\left[\left(x+2\right)\left(3-4x\right)^3\right]\left[\left(2x^2+1\right)^2\right]g(x)=[(x+2)(3−4x)3][(2x2+1)2]
g(x)=[(x+2)(3−4x)3]′[(2x2+1)2]+[(x+2)(3−4x)3][(2x2+1)2]′g\left(x\right)=
\orange{\left[\left(x+2\right)\left(3-4x\right)^3\right]'}
\left[\left(2x^2+1\right)^2\right]
+
\left[\left(x+2\right)\left(3-4x\right)^3\right]
\green{\left[\left(2x^2+1\right)^2\right]'}g(x)=[(x+2)(3−4x)3]′[(2x2+1)2]+[(x+2)(3−4x)3][(2x2+1)2]′

Use product rule again to find the derivative of the first part:
g(x)=[(x+2)′(3−4x)3+(x+2)((3−4x)3)′][(2x2+1)2]+[(x+2)(3−4x)3][2(2x2+1)×(4x)]g\left(x\right)=
\orange{\left[\red{\left(x+2\right)'}\left(3-4x\right)^3+\left(x+2\right)\blue{\left(\left(3-4x\right)^3\right)'}\right]}
\left[\left(2x^2+1\right)^2\right]
+
\left[\left(x+2\right)\left(3-4x\right)^3\right]
\green{\left[2\left(2x^2+1\right)\times\left(4x\right)\right]}g(x)=[(x+2)′(3−4x)3+(x+2)((3−4x)3)′][(2x2+1)2]+[(x+2)(3−4x)3][2(2x2+1)×(4x)]
g(x)=[(1)(3−4x)3+(x+2)(3(3−4x)2×−4)][(2x2+1)2]+[(x+2)(3−4x)3][2(2x2+1)×(4x)]g\left(x\right)=
\orange{\left[\red{\left(1\right)}\left(3-4x\right)^3+\left(x+2\right)
\blue{\left(3\left(3-4x\right)^2\times-4\right)}\right]}
\left[\left(2x^2+1\right)^2\right]
+
\left[\left(x+2\right)\left(3-4x\right)^3\right]\green{\left[2\left(2x^2+1\right)\times\left(4x\right)\right]}g(x)=[(1)(3−4x)3+(x+2)(3(3−4x)2×−4)][(2x2+1)2]+[(x+2)(3−4x)3][2(2x2+1)×(4x)]

We can simplify this a bit:
g(x)=[(3−4x)3−12(x+2)(3−4x)2][(2x2+1)2]+8x(x+2)(3−4x)3(2x2+1)g\left(x\right)=\left[\left(3-4x\right)^3-12\left(x+2\right)\left(3-4x\right)^2\right]\left[\left(2x^2+1\right)^2\right]+8x\left(x+2\right)\left(3-4x\right)^3\left(2x^2+1\right)g(x)=[(3−4x)3−12(x+2)(3−4x)2][(2x2+1)2]+8x(x+2)(3−4x)3(2x2+1)
g(x)=[(3−4x)2(3−4x−12(x+2))][(2x2+1)2]+8x(x+2)(3−4x)3(2x2+1)g\left(x\right)=\left[\left(3-4x\right)^2\left(3-4x-12\left(x+2\right)\right)\right]\left[\left(2x^2+1\right)^2\right]+8x\left(x+2\right)\left(3-4x\right)^3\left(2x^2+1\right)g(x)=[(3−4x)2(3−4x−12(x+2))][(2x2+1)2]+8x(x+2)(3−4x)3(2x2+1)
g(x)=(3−4x)2(−16x−21)(2x2+1)2+8x(x+2)(3−4x)3(2x2+1)g\left(x\right)=\left(3-4x\right)^2\left(-16x-21\right)\left(2x^2+1\right)^2+8x\left(x+2\right)\left(3-4x\right)^3\left(2x^2+1\right)g(x)=(3−4x)2(−16x−21)(2x2+1)2+8x(x+2)(3−4x)3(2x2+1)
g(x)=(3−4x)2(2x2+1)[(−16x−21)(2x2+1)+8x(x+2)(3−4x)]g\left(x\right)=\left(3-4x\right)^2\left(2x^2+1\right)\left[\left(-16x-21\right)\left(2x^2+1\right)+8x\left(x+2\right)\left(3-4x\right)\right]g(x)=(3−4x)2(2x2+1)[(−16x−21)(2x2+1)+8x(x+2)(3−4x)]

Practice: Product Rule for 3 Functions
Find the slope of the tangent line to the curve y=(2x2+1)(x−1)(3x+1)4y=\left(2x^2+1\right)\left(x-1\right)\left(3x+1\right)^4y=(2x2+1)(x−1)(3x+1)4 at the point (1, 0)\left(1,\ 0\right)(1, 0).

Enter the exact value of the slope of the tangent line at this point.

I don't know

Practice: Product Rule
Suppose g(x)g\left(x\right)g(x) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right)f(x)=x2g(x). Given that f′(3)=30f'\left(3\right)=30f′(3)=30, f′′(3)=19f''\left(3\right)=19f′′(3)=19, g(3)=2g\left(3\right)=2g(3)=2, what is g′(3)g'\left(3\right)g′(3) and g′′(3)g''\left(3\right)g′′(3)?

g'(3)=

g''(3)=

I don't know