2.6: Chain Rule

The Chain Rule is more general than the Inverse Function Rule and deals with the case where $x$ and $y$ are both functions of a third variable $t$. Suppose $$x = f(t), \quad y = G(x). \nonumber$$

Thus $x$ depends on $t$, and $y$ depends on $x$. But $y$ is also a function of $t$, $$y = g(t), \nonumber$$ where $g$ is defined by the rule $$g(t) = G(f(t)). \nonumber$$

The function $g$ is sometimes called the composition of $G$ and $f$ (sometimes written $g =G \circ f$).

The composition of $G$ and $f$ may be described in terms of black boxes. The function $g = G \circ f$ is a large black box operating on the input $t$ to produce $g(t) = G(f(t))$. If we look inside this black box (pictured in Figure $\PageIndex{1}$), we see two smaller black boxes, $f$ and $G$. First $f$ operates on the input $t$ to produce $f(t)$, and then $G$ operates on $f(t)$ to produce the final output $g(t) = G(f(t))$.

The Chain Rule expresses the derivative of g in terms of the derivatives of $f$ and $G$. It leads to the powerful method of "change of variables" in computing derivatives
and, later on, integrals.

Figure $\PageIndex{1}$: A composition of two functions.

Chain Rule

Let $f$, $G$ be two real functions and define the new function $g$ by the rule $$g(t) = G(f(t)). \nonumber$$

At any value of $t$ where the derivatives $f'(t)$ and $G'(f(t))$ exist, $g'(t)$ also exists and has the value $$g'(t) = G'(f(t)) f'(t). \nonumber$$

Proof

Let $x = f(t), \quad y = g(t), \quad y = G(x)$.

Take $t$ as the independent variable, and let $\Delta t \neq 0$ be infinitesimal. Form the corresponding increments $\Delta x$ and $\Delta y$. By the Increment Theorem for $x = f(t)$, $\Delta x$ is infinitesimal. Using the Increment Theorem again but this time for $y = G(x)$, we have $$\Delta y = G'(x) \Delta x + \varepsilon \Delta x \nonumber$$

for some infinitesimal $\varepsilon$. Dividing by $\Delta t$, $$\frac{\Delta y}{\Delta t} = G'(x) \frac{\Delta x}{\Delta t} + \varepsilon \frac{\Delta x}{\Delta t}. \nonumber$$

Then taking standard parts, $$\begin{align*} st \left(\frac{\Delta y}{\Delta t}\right) &= G'(x) st \left(\frac{\Delta x}{\Delta t}\right) + 0, \\ g'(t) &= G'(x) f'(t) = G'(f(t)) f'(t). \end{align*}$$

Example $\PageIndex{1}$

Find the derivative of $g(t) = \ln (\sin t)$.

Solution

$g(t)$ is the natural logarithm of the sine of $t$. It can be written in the form $g(t) = G(f(t))$ where $$f(t) = \sin t, \quad G(x) = \ln x. \nonumber$$

We have $$f'(t) = \cos t, \quad G'(s) = \frac{1}{x}. \nonumber$$

By the Chain Rule, $$\begin{align*} g'(t) &= G'(f(t)) f'(t) \\ &= \frac{1}{\sin t} \cdot \cos t = \frac{\cos t}{\sin t}. \end{align*}$$

Example $\PageIndex{2}$

Find the derivative of $g(t) = \sqrt{3t + 1}$. 

Solution

$g(t)$ has the form $g(t) = G(f(t))$, where $$f(t) = 3t + 1, \quad G(x) = \sqrt{x}. \nonumber$$

We have $$f'(t) = 3, \quad G'(x) = \frac{1}{2} x^{-1/2}. \nonumber$$

Then $$\begin{align*} g'(t) &= G'(f(t)) f'(t) \\ \frac{1}{2} (3t + 1)^{-1/2} \cdot 3 = \frac{3}{2 \sqrt{3t + 1}}. \end{align*}$$

In practice it is more convenient to use the Chain Rule with dependent variables $x$ and $y$ instead of functions $f$ and $g$.

Chain Rule with Dependent Variables

Let $$x = f(t), \quad y = g(t) = G(x). \nonumber$$

Assume $g'(t)$ and $G'(x)$ exist. Then $$(\text{i}) \quad \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} \quad\quad (\text{ii}) \quad dy = \frac{dy}{dx} dt \nonumber$$

where $dx/dt$, $dy/dt$ are computed with $t$ as the independent variable, and $dy/dx$ is computed with $x$ as the independent variable.

Let us work Examples $\PageIndex{1}$ and $\PageIndex{2}$ again with dependent variables.

Example $\PageIndex{1}$ continued

Let $x = \sin t, \quad y = \ln x$.

Find $dy/dt$ using Chain Rule $(\text{i})$ and $dy$ by using Chain Rule $(\text{ii})$. 

$$\begin{align*} &(\text{i}) \quad\quad\quad &&\frac{dx}{dt} = \cos t, \quad \frac{dy}{dx} = \frac{1}{x}, \\[4pt] &&&\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{1}{x} \cdot \cos t = \frac{\cos t}{\sin t}. \\[4pt] &(\text{ii}) \quad\quad\quad &&dx = \cos t \ dt, \quad \frac{dy}{dx} = \frac{1}{x}, \\[4pt] &&& dy = \frac{1}{x} \ dx = \frac{1}{x} \cos t \ dt = \frac{\cos t}{\sin t} \ dt. \end{align*}$$

Example $\PageIndex{2}$ continued

Let $x = 3t + 1, \quad y = \sqrt{x}$.

$$\begin{align*} &(\text{i}) \quad\quad\quad &&\frac{dx}{dt} 3, \quad \frac{dy}{dx} = \frac{1}{2} x^{-1/2}, \\[4pt] &&&\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} = \frac{3}{2} x^{-1/2} = \frac{3}{2} (3t + 1)^{-1/2}. \\[4pt] &(\text{ii}) \quad\quad\quad &&dx = 3 \ dt, \quad \frac{dy}{dx} = \frac{1}{2} x^{-1/2}, \\[4pt] &&& dy = \frac{1}{2} x^{-1/2} \ dx = \frac{1}{2} (3t + 1)^{-1/2} \cdot 3 \ dt = \frac{3}{2} (3t + 1)^{-1/2} \ dt. \end{align*}$$

The equation $$\frac{dy}{dt} = \frac{dy}{dt} \frac{dx}{dt} \nonumber$$

with $t$ as the independent variable is trivial. We simply cancel the $dx$ instances. But when $dy/dx$ is computed with $x$ as the independent variable while $dx/dt$ is computed with $t$ as the independent variable, each $dx$ has a different meaning, and the equation is not trivial.

Similarly, the equation $$dy = \frac{dy}{dx} dx \nonumber$$ is trivial with $x$ as the independent variable but not when $t$ is the independent variable in $dy$ and $dx$, while $x$ is independent in $dy/dx$.

The Chain Rule shows that when we change independent variables the equations  $$\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}, \quad dy = \frac{dy}{dx} \ dx \nonumber$$ remain true.

The Inverse Function Rule can be proved from the Chain Rule as follows. Let $$y = f(x), \quad x = g(y) \nonumber$$ be inverse functions whose derivatives exist. Then $$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dy} = 1, \nonumber$$

whence $$\frac{dy}{dx} = \frac{1}{dx/dy}, \quad f'(x) = \frac{1}{g'(y)}. \nonumber$$

Using the Chain Rule we may write the Power Rule in a general form.

Power Rule

Let $r$ be a rational number, and let $u$ depend on $x$. If $u > 0$ and $du/dx$ exists, then $$\frac{d \left(u^{r}\right)}{dx} = ru^{r-1} \frac{du}{dx}. \nonumber$$

This is proved by letting $y = u^{r}$ and computing $dy/dx$ by the Chain Rule. $$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = ru^{r-1} \frac{du}{dx}. \nonumber$$

The Chain Rule has two types of applications.

(1) Given $x = f(t)$ and $y = G(x)$, find $\dfrac{dy}{dt}$. Use $\dfrac{dy}{dt} = \dfrac{dy}{dx} \dfrac{dx}{dt}$.

(2) Given $x = f(t)$ and $y = g(t)$, find $\dfrac{dy}{dx}$. Use $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.

Applications of type (1) often arise when a new dependent variable $x$ is introduced to help compute $\frac{dy}{dt}$. Applications of type (2) arise when two variables $x$ and $y$ both depend on a third variable $t$, for example, when $x$ and $y$ are the coordinates of a moving particle and $t$ is time.

We give three examples of type (1) and then three of type (2).

Example $\PageIndex{3}$

Suppose that by investing $t$ dollars a company can produce $$x = \frac{t}{10} - 100, \quad t \geq 1000, \nonumber$$

items, and that it can sell $x$ items for a total profit of $$y = 5x - \frac{x^{2}}{100}. \nonumber$$

Find $\dfrac{dy}{dt}$, which is the marginal profit with respect to the amount invested.

Solution

We have $$\frac{dx}{dt} = \frac{1}{10}, \quad \frac{dy}{dx} = 5 - \frac{x}{50}. \nonumber$$

By the Chain Rule, $$\begin{align*} \frac{dy}{dt} &= \frac{dy}{dx} \frac{dx}{dt} = \left(5 - \frac{x}{50}\right) \frac{1}{10} \\ &= \left(5 - \frac{\frac{t}{10} - 100}{50}\right) \frac{1}{10} \\ &= 0.7 - \frac{t}{5000}. \end{align*}$$

Thus after $t$ dollars have been invested, an additional dollar will bring $0.7 - t/5000$ dollars of additional profit.

Example $\PageIndex{4}$

Find $dy/dt$ where $y = \left(5t^{2} - 2\right)^{1/4}$.

Solution

Let $$x = 5t^{2} - 2, \quad y = x^{1/4}. \nonumber$$

Then $$\begin{align*} \frac{dx}{dt} &= 10t, \quad \frac{dy}{dx} = \frac{1}{4} x^{-3/4}, \\[4pt] \frac{dy}{dt} &= \frac{dy}{dx} \frac{dx}{dt} = \left(\frac{1}{4} x^{-3/4}\right) (10t) \\ &= \frac{10}{4} \left(5t^{2} - 2\right)^{-3/4} t. \end{align*}$$

Example $\PageIndex{5}$

Find $dy/dx$ where $y = \sqrt{\sin (4x+1) + \cos (4x-1)}$. 

Solution

This problem requires three uses of the Chain Rule. Let $$u = \sin (4x+1) + \cos (4x-1), \quad y = \sqrt{u}. \nonumber$$

Then by the Chain Rule, $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}. \nonumber$$

Now let $$v = \sin (4x+1), \quad w = \cos (4x-1), \quad u = v + w. \nonumber$$

Then $$\frac{du}{dx} = \frac{dv}{dx} + \frac{dw}{dx}. \nonumber$$

We use the Chain Rule twice more to find $dv/dx$ and $dw/dx$. $$\begin{align*} v &= \sin (4x+1). \\[4pt] \frac{dv}{dx} &= \cos (4x+1) \frac{d (4x+1)}{dx} = 4 \cos (4x+1). \\ w &= \cos (4x-1). \\ \frac{dw}{dx} &= -\sin (4x-1) \frac{d (4x-1)}{dx} = -4 \sin (4x-1). \end{align*}$$

Finally, we combine everything to get $$\begin{align*} \frac{dy}{dx} &= \frac{1}{2 \sqrt{u}} \cdot \frac{du}{dx} = \frac{1}{2 \sqrt{u}} \left(\frac{dv}{dx} + \frac{dw}{dx}\right) \\ &= \frac{4 \cos (4x+1) - 4 \sin (4x-1)}{2 \sqrt{\sin (4x+1) + \cos (4x-1)}}. \end{align*}$$  

If a particle is moving in the plane, its position $(x, y)$ at time $t$ will be given by a pair of equations $$x = f(t), \quad y = g(t). \nonumber$$

These are called parametric equations. The slope of the curve traced out by this particle can be found by the Chain Rule, $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}, \nonumber$$ whenever the derivatives exist and $f'(t) \neq 0$. This is a Chain Rule application of type (2).

Example $\PageIndex{6}$

A ball thrown horizontally from a 100-foot cliff at a velocity of 50 ft/sec will follow the parametric equations $$x = 50t, \quad y = 100 - 16t^{2}, \quad \text{in feet}. \nonumber$$ Find the slope of its path at time $t$ (Figure $\PageIndex{2}$).

Solution

$$\frac{dx}{dt} = 50, \quad \frac{dy}{dt} = -32t, \nonumber$$

so $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - \frac{32t}{50}. \nonumber$$

Figure $\PageIndex{2}$: Graph of the ball's path in time.

Example $\PageIndex{7}$

A particle moves according to the parametric equations $$x = t^{3} - t, \quad y = t^{2}. \nonumber$$ Find the slope of its path.

Solution

$$\frac{dx}{dt} = 3t^{2} - 1, \quad \frac{dy}{dt} = 2t. \nonumber$$

Therefore, $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{3t^{2} - 1}, \quad t \neq \pm \sqrt{1/3}. \nonumber$$

We see from Figure $\PageIndex{3}$ that the path of this particle is not the graph of a function, and in fact contains a loop and crosses the point $(0, 1)$ twice, at $t = -1$ and $t = 1$. The path is vertical at the points $t = \pm \sqrt{1/3}$, where there is no slope. At the point $(0, 1)$, the two slopes of the path are $dy/dx = -1$ when $t = -1$, and $dy/dx = 1$ when $t = 1$.

Figure $\PageIndex{3}$: Graphs of the particle's path in time.

Example $\PageIndex{8}$

A particle moving according to the parametric equations $$x = \cos t, \quad y = \sin t \nonumber$$ will move counterclockwise around the unit circle at one radian per second beginning at the point $(1, 0)$, as shown in Figure $\PageIndex{4}$. Find the slope of its path at time $t$.

Figure $\PageIndex{4}$: Path in time of a particle moving around the unit circle.

Solution

$$\frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = \cos t. \nonumber$$

The slope is $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - \frac{\cos t}{\sin t}. \nonumber$$

In terms of $x$ and $y$ the slope is $$\frac{dy}{dx} = - \frac{x}{y}. \nonumber$$

Problems for Section 2.6

In Problems 1-44, find $dy/dx$.I 

1.	$y = \sqrt{x + 2}$	2.	$y = \sqrt{7 + 4x}$
3.	$y = \sqrt{5 - x}$	4.	$y = \sqrt{1 - 10x}$
5.	$y = \dfrac{1}{\sqrt{2 + 3x}}$	6.	$y = \dfrac{1}{\sqrt{4 - x}}$
7.	$y = \sqrt[3]{6x + 1}$	8.	$y = \sqrt[5]{2 - 3x}$
9.	$y = \sqrt{x^{2} + 1}$	10.	$y = \sqrt{1 - x^{2}}$
11.	$y = \sin (3x)$	12.	$y = \cos (4-2x)$
13.	$y = \sin \left(x^{-2}\right)$	14.	$y = \cos \sqrt{x}$
15.	$y = e^{4x}$	16.	$y = e^{-x^{2}}$
17.	$y = e^{\cos x}$	18.	$y = \ln (\ln x)$
19.	$y = \cos u, \quad u = e^{x}$	20.	$y = \tan u, \quad u = \ln x$
21.	$y = u^{10}, \quad 1 - 4x$	22.	$y = u^{-10}, \quad 1 - x^{2}$
23.	$y = \sin u + \sin v, \quad u = 1 - x^{2}, \quad v = 2x - 1$
24.	$y = e^{u} + e^{v}, \quad 1 - 3x, \quad v = 3 - 4x$
25.	$y = e^{u}, \quad u = \sqrt{v}, \quad v = \sin x$
26.	$y = \ln u, \quad u = \tan v, \quad v = 1/x$
27.	$y = u^{-1/3}, \quad u = 1 + \sqrt{v}, \quad v = x^{2} - 1$
28.	$y = u^{-1}, \quad u = 3v + 4, \quad v = 1/(x + 1)$
29.	$y = u^{4}, \quad u = 1 + 1/v, \quad v = x^{3} + 1$
30.	$y = u^{2} + 1, \quad u = v^{2} + 1, \quad v = x^{2} + 1$
31.	$y = \left(\sqrt{x^{2} + 1} + \sqrt{x^{2} + 1}\right)^{1/3}$	32.	$y = \left(x + \sqrt{3 - 4x}\right)^{-1/2}$
33.	$y = 3x \sin (2x - 1)$	34.	$y = \sin (2x) \cos (3x)$
35.	$x = \cos (3t), \quad y = \sin (3t)$	36.	$x = e^{t}, \quad y = \ln t$
37.	$x = \sin t, \quad y = \sin (2t)$	38.	$x = \sin (e^{t}), \quad y = \cos (e^{t})$
39.	$x = \ln (t + 1), \quad y = t^{2}$	40.	$x = e^{\cos t}, \quad y = e^{\sin t}$
41.	$x = \sqrt{t^{2} - 4}, \quad y = t^{2}$	42.	$x = 1 + \sqrt[3]{t}, \quad y = 2 + \sqrt[3]{t}$
43.	$x = \sqrt[3]{t + 1}, \quad y = \sqrt[3]{t + 2}$	44.	$x = \dfrac{2t + 1}{t + 2}, \quad y = \dfrac{2t + 3}{t + 2}$
45.	A particle moves in the plane according to the parametric equations $$x = t^{2} + 1, \quad y = 3t^{3}. \nonumber$$ Find the slope of its path.
46.	An ant moves in the plane according to the equations $$x = \left(1 - t^{2}\right)^{-1}, \quad y = \sqrt{t}. \nonumber$$ Find the slope of its path.
$\square$ 47.	Let $y$ depend on $u$, $u$ depend on $v$, and $v$ depend on $x$. Assume the derivatives $dy/du$, $du/dv$, and $dv/dx$ exist. Prove that $$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dv} \frac{dv}{dx}. \nonumber$$
$\square$ 48.	Let the function $f(x)$ be differentiable for all $x$, and let $g(x) = f(f(x))$. Show that $g'(x) = f'(f(x)) f'(x)$.