Lesson Part 1

Recall

• We have learned how to differentiate a polynomial function expressed in its fully expanded and simplified form. For example, \(f(x)=ax^2+bx+c\).

In This Module

• We will learn how to differentiate a polynomial in factored form. For example, \(f(x) = (x^2 + 2)(x + 5)\).
• We will determine a method of finding the derivative of \(f(x)=p(x)q(x)\), which does not require us to expand and simplify before differentiating.

Product Rule of Two Functions

Let's use first principles to find a general expression for the derivative of the product of two differentiable functions, \(p(x)\) and \(q(x)\).

Let \( f(x) = p(x)q(x) \).

Then, by the definition of the derivative,

\[ \begin{align*} f'(x) &=\displaystyle \lim_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} \\ &= \displaystyle \lim_{h \rightarrow 0} \dfrac{p(x+h)q(x+h) - p(x)q(x)}{h} \end{align*} \]

When we look at the two terms in the numerator, there is not much that can be done. We cannot simplify further, nor can we factor this expression.

Now, consider the expression \(- p(x)q(x+h) + p(x)q(x+h)\).

Here, these two terms are equivalent except for their sign. This expression equals zero, so we can insert this expression into the limit calculation without changing its value.

\[f'(x) = \displaystyle \lim_{h \rightarrow 0} \dfrac{p(x+h)q(x+h) - p(x)q(x+h) + p(x)q(x+h) - p(x)q(x)}{h} \]

Look closely at the first two terms in the numerator. Quantity \(q(x + h)\) is common to both. Therefore, we can remove this as a common factor. In the last two terms, \(p(x)\) is common to both and can be, again, removed as a common factor.

\[f'(x) = \displaystyle \lim_{h \rightarrow 0} \dfrac{q(x+h) [p(x+h) - p(x)]+ p(x)[q(x+h) - q(x)]}{h} \]

Once we have removed these common factors, we really have two terms left in the numerator. Once the common factors have been removed, we can now break this expression into two separate fractions.

\[f'(x) = \displaystyle \lim_{h \rightarrow 0} \left[\dfrac{q(x+h)[p(x+h) - p(x)]}{h} + \dfrac{p(x)[q(x+h) - q(x)]}{h}\right] \]

By breaking the expression into two separate functions, we can apply the laws of limits.

\[f'(x) = \displaystyle \lim_{h \rightarrow 0}\left[q(x+h)\left(\dfrac{p(x+h) - p(x)}{h}\right) + p(x)\left(\dfrac{q(x+h) - q(x)}{h}\right)\right]\]

Assuming that each limit in the following sums exist, we can apply the properties of limits to simplify the following,

by limit property 3

\[f'(x) = \displaystyle \lim_{h \rightarrow 0} \left[q(x+h)\left(\dfrac{p(x+h) - p(x)}{h}\right)\right] + \lim_{h \rightarrow 0}\left[p(x)\left(\dfrac{q(x+h) - q(x)}{h}\right)\right]\]

by limit property 5

\[f'(x) = \left[\displaystyle\lim_{h \rightarrow 0}q(x+h)\right] \underbrace{\left[\displaystyle\lim_{h \rightarrow 0} \dfrac{p(x+h) - p(x)}{h}\right]}_{p'(x)}+ \left[\displaystyle\lim_{h \rightarrow 0}p(x)\right] \underbrace{\left[\displaystyle\lim_{h \rightarrow 0} \dfrac{q(x+h) - q(x)}{h}\right]}_{q'(x)}\]

This breaks our full expression into four separate limits.

From left to right, look first at the second limit. This should be recognizable to you. We have \(\dfrac{p(x + h) - p(x)}{h}\), and we're finding \(\displaystyle\lim_{h \rightarrow 0}\). So really, we're asked to find \(p'(x)\).

Now look at the last limit. We have \(\dfrac{q(x+h) - q(x)}{h}\), and the \(\displaystyle\lim_{h \rightarrow 0}\). Therefore, this is also \(q'(x)\).

We can further simplify by substituting \(0\) in for \(h\) into the first and third limit, and we get the expression

\[f'(x) = \displaystyle q(x+0) \underbrace{\left[\displaystyle\lim_{h \rightarrow 0} \dfrac{p(x+h) - p(x)}{h}\right]}_{p'(x)} + \displaystyle p(x) \underbrace{\left[\displaystyle\lim_{h \rightarrow 0} \dfrac{q(x+h) - q(x)}{h}\right]}_{q'(x)}\]

since we are assuming that \(p'(x)\) and \(q'(x)\) exist

\[f'(x) = p'(x)q(x+0) + p(x)q'(x) \]

Therefore, the expression for product rule simplifies to

\[f'(x)= p'(x)q(x) + p(x)q'(x)\]

A helpful way to remember the product rule is to think the derivative of the first expression times the second expression plus the first expression times the derivative of the second expression.

Formally, the product rule is defined as follows.

The Product Rule

If \( f(x) = p(x)q(x) \), where \( p(x) \) and \( q(x) \) are differentiable functions, then

\[ f'(x) = p'(x)q(x) + p(x)q'(x) \]

In Leibniz notation,

\[\dfrac{d}{dx}(uv)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\]

where \(u(x)\) and \(v(x)\) are differentiable functions.

Note:

When differentiating expressions, we must always consider the domain on which the expression is differentiable. If \(p(x)\) or \(q(x)\) are not differentiable at the point \(a\), then \(f(x)\) may not be differentiable at that point.

For example, suppose \(f(x) = x^2\lvert x\rvert\). Here, \(q(x)=\lvert x\rvert\) and \(q(x)\) has a cusp at \(x=0\), so \(q(x)\) is not differentiable at \(x=0\). Therefore, \(f(x)\) is also not differentiable at \(x=0\).

When differentiating expressions, we must always consider the domain at which the expression is differentiable.

Lesson Part 2

Examples

Let's now try example 1.

Example 1

Without expanding, differentiate \( h(x) = 4x^2(2x^3 - 4x + 9) \).

Solution

In this example, let \(p(x) = 4x^2\) and \(q(x) = 2x^3 - 4x + 9 \).

Using previously learned differentiation rules, we know that \( p'(x) = 8x \) and \( q'(x)= 6x^2 -4 \).

Now substitute \( p(x), q(x), p'(x),\) and \(q'(x) \) into the formula for product rule to give

\[ \begin{align*} f'(x) &= p'(x)q(x) + p(x)q'(x) \\ &= (8x)(2x^3 - 4x + 9) + 4x^2(6x^2 - 4) \\ \end{align*} \]

Once we have substituted these expressions into the formula for product rule, we will notice that there is a common factor in both of these two terms. First of all, we can remove a common factor of \(4\). And we can also remove a common factor of \(x\).

\[ f'(x) = 4x[2(2x^3 - 4x + 9) + x(6x^2 - 4)] \]

Once \(4x\) is removed from these two terms, we can expand and simplify the remaining parts, giving us the derivative of \(f(x)\).

\[ \begin{align*} f'(x) &= 4x(4x^3 - 8x + 18 + 6x^3 - 4x) \\ &= 4x(10x^3 - 12x + 18) \end{align*} \]

Example 2

Differentiate \( f(x) = (\sqrt{x} + 1)(3x^2 - 5x) \) using the product rule.

Solution

Our first factor is \(\sqrt{x} + 1\).

Let \(p(x) = \sqrt{x} + 1\)

To differentiate this expression, we're going to first express \(x\) not with a radical sign but instead with a rational exponent.

\[p(x) = x^{\frac{1}{2}}+1\]

Differentiating, we have

\[ \begin{align*} p'(x) &= \dfrac{1}{2}x^{-\frac{1}{2}} \\ &=\dfrac{1}{2\sqrt{x}}, \quad \text{for} \quad x \gt 0 \end{align*} \]

The second factor in our expression of \(f(x)\) is \(3x^2 - 5x\).

Let \( q(x) = 3x^2 - 5x \).

Differentiating, we have \( q'(x) = 6x - 5 \).

We are now ready to substitute \(p(x)\), \(p'(x)\), \(q(x)\), and \(q'(x)\) into the formula for product rule.

\[ \begin{align*} f'(x) &= p'(x)q(x) + p(x)q'(x) \\ &= \left(\dfrac{1}{2}x^{-\frac{1}{2}}\right)(3x^2 - 5x) + (x^{\frac{1}{2}}+1)(6x - 5) \end{align*} \]

Note that, in this case, we're going to continue to use rational exponents rather than radical signs on our terms of \(x\). This will make simplifying much easier.

Expand and simplify each of the terms in this expression, and then convert the rational exponents on \(x\) back into their radical form.

\[ \begin{align*} f'(x) &= \dfrac{3}{2}x^{\frac{3}{2}} - \dfrac{5}{2}x^{\frac{1}{2}} + 6x^{\frac{3}{2}} - 5x^{\frac{1}{2}} + 6x - 5 \\ &= \dfrac{15}{2} \sqrt{x^3} + 6x - \dfrac{15}{2} \sqrt{x} - 5 \end{align*} \]

The function \(f(x)\) has domain \(x\geq 0\), so we can say that it has a right-side derivative at \(x=0\), namely

\[f'(0^+)=-5\]

But it does not have a left-side derivative, since this function is not defined for \(x \lt 0\).

In other words, we cannot approach the value \(x=0\) from the left-side, so the derivative does not exist at \(x=0\).

Lesson Part 3

What would happen if we had the product of three or more functions? Would product rule apply in this case?

Extension: Product of Three or More Functions

Even though the product rule is stated for the product of two functions, it can be applied recursively (multiple times) on functions that are products of three or more functions.

For example, here is a derivation of the general derivative of a function

\[ f(x) = p(x)q(x)r(x) \]

where \(p, q,\) and \(r\) are differentiable functions.

If we were to apply the product rule as we know it to this point, we need two functions. So in this case, we will write \(f(x)\) as the product

\[f(x)= [p(x)][q(x)r(x)]\]

Once \(q(x)\) and \(r(x)\) are grouped together, we can differentiate this expression using the product rule as we know it up to this point.

Applying the product rule,

\[f'(x) = p'(x)[q(x)r(x)] + p(x)[q(x)r(x)]' \]

But notice how we have still not differentiated \(q(x)r(x)\). Now we will apply the product rule again to only this expression.

\[ f'(x) = p'(x)q(x)r(x) + p(x)[q'(x)r(x) + q(x)r'(x)] \]

We are now going to expand and simplify the terms. And we get the formula for the extended product rule.

\[ f'(x) = p'(x)q(x)r(x) + p(x)q'(x)r(x) + p(x)q(x)r'(x) \]

Note:
There is a nice symmetry in the general formula for the derivative of a product of three functions.
A similar symmetry appears for the derivative of products of four or more functions.

Examples

Let's now apply the extended product rule for three functions.

Example 3

Differentiate \( y = x(4x-1)(6x + 3) \) using the extended product rule.

Solution

In the first term, we'll differentiate x and multiply by the second two factors. In the second term, we multiply the first and third factors by the derivative of the middle factor. And in the last term, we will multiply the first and second factor by the derivative of the third factor.

\[\dfrac{dy}{dx} = x'(4x-1)(6x+3) + x(4x-1)'(6x+3) + x(4x-1)(6x+3)'\]

Once this is done, expand and simplify each of the terms to receive the derivative.

\[ \begin{align*} \dfrac{dy}{dx} &= (4x-1)(6x+3) + x(4)(6x+3) + x(4x-1)(6) \\ &= 24x^2 + 6x - 3 + 24x^2 + 12x + 24x^2 - 6x \\ &= 72x^2 + 12x - 3 \end{align*} \]

It is worth noting that a simpler approach to differentiating this expression may have been to simply expand and simplify the original expression before differentiating. However, it was nice to try to use the extended product rule for this function.

Lesson Part 4

Examples

Let's now apply the product rule to this challenging question.

Challenge Question

The tangent line to the curve of \(y=f(x)\) at \(x=2\) is \(y=4x+1\).

The tangent line to the curve of \(y=g(x)\) at \(x=2\) is \(y=3x-2\).

Find the equation of the tangent line to the curve \(y=f(x)g(x)\) at \(x=2\).

Solution

Since the tangent line to the curve of \(y=f(x)\) at \(x=2\) is \(y=4x+1\), we know that at \(x=2\), \(m_{\text{tangent}}=4,\) and thus \(f'(2)=4\).

Also, by substituting \(x=2\) into the equation of the tangent line, we can find the point of tangency.

When \(x=2\), \(y =4x+1=4(2)+1=9\).

Therefore, the point of tangency is \((2,9)\) which tells us that \((2, f(2))=(2,9)\).

Similarly, the tangent line to the curve of \(y=g(x)\) at \(x=2\) is \(y=3x-2\).

Therefore, we know that at \(x=2\), \(m_{\text{tangent}}=3,\) and thus \(g'(2)=3\).

Also, when \(x=2\), \(y =3x-2=3(2)-2=4\).

Therefore, the point of tangency is \((2,4)\) which tells us that \((2, g(2))=(2,4)\).

Next, to find the equation of the tangent line to the curve \(y=f(x)g(x)\) at \(x=2\), we will first use the product rule to differentiate this expression to find the slope of this tangent when \(x=2\).

By the product rule, \(y=f(x)g(x)\) has derivative

\[\dfrac{dy}{dx}=f'(x)g(x)+f(x)g'(x).\]

Since we are looking to evaluate this expression when \(x = 2\), substitute \(2\) in the place of \(x\).

Thus, at \(x=2\)

\[ \begin{align*} \left.\dfrac{dy}{dx}~\right\vert_{x=2} &= f'(2)g(2) + f(2)g'(2) \\ &= (4)(4) + (9)(3) \\ &=43 \end{align*} \]

Therefore, \(m_{\text{tangent}}=43\).

Now that we know that the slope of the tangent is \(43\), we need a point on the tangent to determine its equation.

When \(x=2\), \(y=f(2)g(2)=(9)(4)=36\).

Therefore, the point of tangency to \(y=f(x)g(x)\) is \((2,36)\).

Finally, let's use the slope of the tangent line to \(y=f(x)g(x)\) and the point of tangency to determine the equation of the tangent line.

The slope-point equation of the line \(y-y_0=m(x-x_0)\) gives

\[ \begin{align*} y - 36 &= 43(x-2) \\ y - 36 &= 43x - 86 \\ 43x - y - 50 &= 0 \end{align*} \]

Thus, the equation of the tangent line to \(y=f(x)g(x)\) at \(x=2\) is \(43x-y-50=0\).

Quiz

See the quiz in the side navigation.