2.5: Chain Rule

Find the derivative of [latex]y=\left(4x^3+15x\right)^2[/latex]

Answer:

This is not a simple polynomial, so we can’t use the basic building block rules yet. It is a product, so we could write it as [latex]y=\left(4x^3+15x\right)^2=\left(4x^3+15x\right)\left(4x^3+15x\right)[/latex] and then use the product rule. Or we could multiply it out and simply differentiate the resulting polynomial. Let’s try do it the second way:
\[ \begin{align*}
y & = \left(4x^3+15x\right)^2\\
& = 16x^6+120x^4+225x^2\\
y’ & = 96x^5+480x^3+450x
\end{align*} \]

Find the derivative of [latex]y=\left(4x^3+15x\right)^2[/latex]

Answer:

This is the same one we did before by multiplying out. This time, let’s use the Chain Rule: The inside function is what appears inside the parentheses: [latex]4x^3+15x[/latex]. The outside function is the first thing we find as we come in from the outside – it’s the square function, [latex](\text{inside})^2[/latex].

In [latex]y=\left(4x^3+15x\right)^2[/latex], the inside function is [latex]4x^3+15x[/latex], which is then squared, so the square is the outside function.

The derivative of this outside function is [latex](2\cdot\text{inside})[/latex]. Now using the chain rule, the derivative of our original function is [latex](2\cdot\text{inside})[/latex] times the derivative of the inside (which is [latex]12x^2+15[/latex]):
\[ y’=2\left(4x^3+15x\right)\left(12x^2+15 \right)\]

Find the derivative of [latex]y=\left(4x^3+15x\right)^{20}[/latex].

Answer:

Now we have a way to handle this one. It’s the derivative of the outside times the derivative of the inside.

The outside function is [latex]\left(\text{inside}\right)^{20}[/latex], which has derivative [latex]20\left(\text{inside}\right)^{19}[/latex], so \[y’=20\left(4x^3+15x\right)^{19}\left(12x^2+15\right).\]

Differentiate [latex]y=e^{x^2+5}[/latex].

Answer:

This isn’t a simple exponential function because the exponent is not simply the variable – it is another function (a polynomial). As a result, this is a composition of functions.

To determine the inside function and the outside function, analyze what happens to your variable. Here, our variable [latex]x[/latex] is first used in a polynomial, which is [latex]x^2+5[/latex]. This tells us that [latex]x^2+5[/latex] is the inside function. Then the polynomial is put into the exponent of [latex]e[/latex], so [latex]e^{\text{inside}}[/latex] is the outside function.

Now we can use the Chain Rule: We want the derivative of the outside times the derivative of the inside. The outside is the [latex]e[/latex] to the something function, so its derivative is the same thing. The derivative of what’s inside is [latex]2x[/latex]. So \[\frac{d}{dx}\left( e^{x^2+5} \right)= \left( e^{x^2+5} \right)\cdot (2x)\]

The table gives values for [latex]f[/latex] , [latex]f'[/latex] , [latex]g[/latex], and [latex]g'[/latex] at a number of points. Use these values to determine [latex]( f \circ g )(x)[/latex] and [latex]( f \circ g ) '(x)[/latex] at [latex]x = -1[/latex] and 0.

Answer:

[latex](f\circ g)(-1)  = f\left(g(-1)\right)=f(3)=0[/latex]

[latex](f\circ g)(0) = f\left(g(0)\right)=f(1)=1\\[/latex]

[latex](f\circ g)'(-1) = f'\left(g(-1)\right)\cdot g'(-1)=f'(3)\cdot (0)=(2)(0)=0[/latex]

[latex](f\circ g)'(0) = f'\left(g(0)\right)\cdot g'(0)=f'(1)\cdot (2)=(-1)(2)=-2[/latex]

If 2400 people now have a disease, and the number of people with the disease appears to double every 3 years, then the number of people expected to have the disease in [latex]t[/latex] years is [latex]y=2400\cdot 2^{t/3}[/latex].

a. How many people are expected to have the disease in 2 years?

b. When are 50,000 people expected to have the disease?

c. How fast is the number of people with the disease expected to grow now and 2 years from now?

Answer:

a. In 2 years, [latex]y = 2400\cdot 2^{2/3} \approx 3,810[/latex] people.

b. We know [latex]y = 50,000[/latex], and we need to solve [latex]50,000 = 2400\cdot 2^{t/3}[/latex] for [latex]t[/latex].

\[
\begin{align*}
\frac{50000}{2400} & = 2^{t/3} \\
\Rightarrow&\ln\left(\frac{50000}{2400}\right)  = \ln\left(2^{t/3}\right)\\
\Rightarrow&\ln\left(\frac{50000}{2400}\right)  = \frac{t}{3}\cdot \ln(2)\\
\Rightarrow&t = \frac{3\ln\left(\frac{50000}{2400}\right)}{\ln(2)}\approx 13.14\text{ years}
\end{align*}
\]

Therefore we expect 50,000 people to have the disease about 13.14 years from now.

c. This is asking for [latex]\frac{dy}{dt}[/latex] when [latex]t =[/latex] 0 and 2 years. When we analyze [latex]y[/latex], we can see that the exponent of 2 is not simply the variable [latex]t[/latex], but a function of [latex]t[/latex] ([latex]t[/latex] divided by 3). Therefore, the function [latex]y=2^{t/3}[/latex] is a composition of functions and so, to find its derivative, we must use the chain rule. The inside function is [latex]\frac{1}{3}t[/latex] and the outside function is [latex]2^{\text{inside}}[/latex].
\[ \begin{align*}
\frac{dy}{dt} & = \frac{d}{dt}\left(2400\cdot 2^{t/3}\right) \\
& = 2400\cdot 2^{t/3}\cdot \ln(2)\cdot\frac{1}{3} \\
& \approx 554.5\cdot 2^{t/3}
\end{align*} \]
So, at [latex]t=0[/latex] the rate of growth of the disease is approximately [latex]554.5\cdot 2^0 \approx 554.5[/latex] people/year. In 2 years the rate of growth will be approximately [latex]554.5\cdot 2^{2/3} \approx 880[/latex] people/year.

Find [latex]\frac{d}{dx}\left( e^{3x}\cdot\ln(5x+7) \right)[/latex]

Answer: Let’s first analyze the function.

Loking at the whole expression, we see that this is a product of two compositions of functions. So we’ll need the Product Rule first. We’ll fill in the pieces we know, and then we can figure the rest as separate steps and substitute in at the end:

\[\frac{d}{dx}\left( e^{3x}\cdot\ln(5x+7) \right)=\left( \frac{d}{dx}\left( e^{3x}\right)\right)\cdot\ln(5x+7)+ e^{3x}\cdot \left(\frac{d}{dx}\left(\ln(5x+7) \right)\right)\]
Now as separate steps, we’ll find \[\frac{d}{dx}\left( e^{3x}\right)=3e^{3x} \quad \text{ (using the Chain Rule)}\] and \[\frac{d}{dx}\left(\ln(5x+7) \right)=\frac{1}{5x+7}\cdot 5 \quad \text{ (also using the Chain Rule)}.\]
Finally, to substitute these in their places:\[\frac{d}{dx}\left( e^{3x}\cdot\ln(5x+7) \right)=\left( 3e^{3x}\right)\cdot\ln(5x+7)+ e^{3x}\cdot \left(\frac{1}{5x+7}\cdot 5\right)\]
(We can stop here – we don’t need to try to simplify any further.)

Differentiate [latex]z=\left(\dfrac{3t^3}{e^t(t-1)}\right)^4[/latex]

Answer:

Don’t panic! As we look at the function, we can see that we have some complicated thing, raised to the power of 4. That tells us that we are dealing with a composition of functions: to find the value of the function, first the thing in the bracket gets calculated, then it is raised to the power of 4. So the inside function is what is inside the bracket. This tells us that we will have to use the chain rule. In order to use the chain rule, we will have to calculate the derivative of the outside function (power function) and the derivative of the inside function (the complicated thing).

Now, when we look at the inside function, we can see that it is a quotient, or a fraction, of two functions. This means we will have to use the Quotient Rule when we differentiate the inside function. This quotient is made of a polynomial in the numerator and a product of functions in the denominator, for which we will have to use the product rule.

Step One: Use the Chain Rule. The derivative of the outside times the derivative of the inside:

\[
\frac{dz}{dt}=\frac{d}{dt}\left(\frac{3t^3}{e^t(t-1)}\right)^4=4\left(\frac{3t^3}{e^t(t-1)}\right)^3\cdot \frac{d}{dt}\left(\frac{3t^3}{e^t(t-1)}\right)
\]

Now we’re one step inside, and we can concentrate on just the [latex]\frac{d}{dt}\left(\frac{3t^3}{e^t(t-1)}\right)[/latex]. This is the derivative of a quotient of two functions.

Step Two: Use the Quotient Rule. The derivative of the numerator is straightforward, so we can just calculate it. The derivative of the denominator uses product rule, so we’ll leave it for now:

\[
\frac{d}{dt}\left(\frac{3t^3}{e^t(t-1)}\right)=\frac{\left( 9t^2 \right)\left( e^t(t-1) \right)-\left( 3t^3 \right)\left( \frac{d}{dt}\left( e^t(t-1) \right) \right)}{\left(e^t(t-1)\right)^2}
\]

Now we’ve gone one more step inside, and we can concentrate on just the denominator [latex]\frac{d}{dt}\left( e^t(t-1) \right)[/latex], which involves a product.

Step Three: Use the Product Rule: \[ \frac{d}{dt}\left( e^t(t-1)\right) = \left( e^t \right)(t-1)+\left( e^t \right)(1)\]

And now we’re all the way in – no more derivatives to take!

Step Four: Now it’s just a question of substituting back – be careful now!

\[
\frac{d}{dt}\left( e^t(t-1)\right) = \left( e^t \right)(t-1)+\left( e^t \right)(1)
\]

so

\[
\frac{d}{dt}\left(\frac{3t^3}{e^t(t-1)}\right)=\frac{\left( 9t^2 \right)\left( e^t(t-1) \right)-\left( 3t^3 \right)\left( \left( e^t \right)(t-1)+\left( e^t \right)(1) \right)}{\left(e^t(t-1)\right)^2}
\]

so

\[
\begin{align*}
\frac{dz}{dt}&=\frac{d}{dt}\left(\frac{3t^3}{e^t(t-1)}\right)^4\\\\
&=4\left(\frac{3t^3}{e^t(t-1)}\right)^3\cdot \left( \frac{\left( 9t^2 \right)\left( e^t(t-1) \right)-\left( 3t^3 \right)\left( \left( e^t \right)(t-1)+\left( e^t \right)(1) \right)}{\left(e^t(t-1)\right)^2} \right)
\end{align*}
\]

Phew!

Show that [latex]f(x)=\sqrt[3]{x}=x^{1/3}[/latex] is not differentiable at [latex]x = 0[/latex].

Answer:

At [latex]x=0[/latex], [latex]f(x)=0^{1/3}=0[/latex] and so the function is defined at 0.

Finding the derivative, we get [latex]f(x)=\frac{1}{3}x^{-2/3}=\frac{1}{3x^{2/3}}[/latex]. At [latex]x = 0[/latex], the derivative function is undefined.

From the graph, we can see that the tangent line to this curve at [latex]x = 0[/latex] is vertical with undefined slope, which is why the derivative does not exist at [latex]x = 0[/latex].

Show that [latex]f(x)=|x|[/latex] is not differentiable at [latex]x = 0[/latex].

Answer:

On the left side of the graph, the slope of the line is -1. On the right side of the graph, the slope is +1. There is no well-defined tangent line at the sharp corner at [latex]x = 0[/latex], so the function is not differentiable at that point.