2.3 The Chain Rule
Differentiating a composition: the derivative of the outside evaluated at the inside, times the derivative of the inside.
We have techniques for differentiating basic functions and for sums, differences, products, quotients and constant multiples. But they do not let us differentiate compositions such as $h(x)=\sin\left(x^3\right)$ or $k(x)=\sqrt{3x^2+1}$. In this section we study the rule for the derivative of a composition of two or more functions.
📐 Rule — The Chain Rule
Let $f$ and $g$ be functions. For all $x$ in the domain of $g$ for which $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, the derivative of the composite function
$$y=(f\circ g)(x)=f\big(g(x)\big)$$
is given by
$$\frac{dy}{dx}=f'\big(g(x)\big)\cdot g'(x).$$
Alternatively, if $y$ is a function of $u$ and $u$ is a function of $x$, then
$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}.$$
📐 Chain rule for a composition of three functions
For all values of $x$ for which the function is differentiable, if $y=f\big(g(h(x))\big)$ then
$$\frac{dy}{dx}=f'\big(g(h(x))\big)\cdot g'\big(h(x)\big)\cdot h'(x).$$
Peel the layers from the outside in, and multiply as you go.
The Chain and Power Rules Combined
To differentiate $y=\big(g(x)\big)^n$, think of $y=f(g(x))$ with $f(x)=x^n$, so that $f'\big(g(x)\big)=n\big(g(x)\big)^{n-1}$.
📐 Rule — Power rule for a composition of functions
For all values of $x$ for which the derivative is defined,
$$\frac{d}{dx}\left[\big(g(x)\big)^n\right]=n\big(g(x)\big)^{n-1}\cdot g'(x).$$
📐 Theorem 2.14 — Chain rule with trigonometric, exponential and logarithmic functions
sine$\dfrac{d}{dx}\big[\sin(g(x))\big]=\cos(g(x))\cdot g'(x)$
cosine$\dfrac{d}{dx}\big[\cos(g(x))\big]=-\sin(g(x))\cdot g'(x)$
tangent$\dfrac{d}{dx}\big[\tan(g(x))\big]=\sec^2(g(x))\cdot g'(x)$
cotangent$\dfrac{d}{dx}\big[\cot(g(x))\big]=-\csc^2(g(x))\cdot g'(x)$
secant$\dfrac{d}{dx}\big[\sec(g(x))\big]=\sec(g(x))\tan(g(x))\cdot g'(x)$
cosecant$\dfrac{d}{dx}\big[\csc(g(x))\big]=-\csc(g(x))\cot(g(x))\cdot g'(x)$
exponential$\dfrac{d}{dx}\left[e^{g(x)}\right]=e^{g(x)}\cdot g'(x)$
logarithm$\dfrac{d}{dx}\big[\ln g(x)\big]=\dfrac{1}{g(x)}\cdot g'(x)$
✏️ Example 1 — Using the chain and power rules
Find the derivative of $h(x)=\dfrac{1}{\left(3x^2+1\right)^2}$.
💡 Show solution▶
Step 1 Rewrite with a negative exponent so the power rule applies:
$$h(x)=\frac{1}{\left(3x^2+1\right)^2}=\left(3x^2+1\right)^{-2}.$$
Step 2 Apply the power rule for compositions with $g(x)=3x^2+1$, so $g'(x)=6x$:
$$h'(x)=-2\left(3x^2+1\right)^{-3}\cdot(6x).$$
Step 3 Rewrite in the original form:
$$h'(x)=\frac{-12x}{\left(3x^2+1\right)^3}.$$
✏️ Example 2 — Chain and power rules with a trigonometric function
Find the derivative of $y=\sin^3x$.
💡 Show solution▶
Step 1 Recall that $\sin^3x=(\sin x)^3$, so the outer function is the cube and the inner function is $g(x)=\sin x$.
Step 2 Apply the power rule for compositions:
$$\frac{dy}{dx}=3(\sin x)^2\cdot\cos x=3\sin^2x\cos x.$$
Warning $\sin^3 x$ and $\sin\left(x^3\right)$ are different functions with different derivatives. Read the notation carefully before choosing the inner function.
✏️ Example 3 — Using the chain rule on a trigonometric function
Find the derivative of $f(x)=\sec\left(4x^5+2x\right)$.
💡 Show solution▶
Step 1 Take $g(x)=4x^5+2x$ and apply the chain rule to $f(x)=\sec(g(x))$:
$$f'(x)=\sec\left(4x^5+2x\right)\tan\left(4x^5+2x\right)\cdot\left(20x^4+2\right).$$
Step 2 Write the factor from the inside first, as is customary:
$$f'(x)=\left(20x^4+2\right)\sec\left(4x^5+2x\right)\tan\left(4x^5+2x\right).$$
✏️ Example 4 — Differentiating a composite of three functions
Find the derivative of $y=\cos^4\left(7x^2+1\right)$.
💡 Show solution▶
Step 1 Rewrite to make the three layers visible:
$$y=\left[\cos\left(7x^2+1\right)\right]^4.$$
The layers are: fourth power $\circ$ cosine $\circ$ the polynomial $7x^2+1$.
Step 2 Apply the chain rule for three functions, peeling from the outside:
$$\frac{dy}{dx}=4\left[\cos\left(7x^2+1\right)\right]^3\cdot\left(-\sin\left(7x^2+1\right)\right)\cdot(14x).$$
Step 3 Collect the constants:
$$\frac{dy}{dx}=-56x\cos^3\left(7x^2+1\right)\sin\left(7x^2+1\right).$$
Leibniz form In Leibniz notation the same computation reads $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dv}\cdot\dfrac{dv}{dx}$, with $y=u^4$, $u=\cos v$ and $v=7x^2+1$.
Exercises
I
Given $y=f(u)$ and $u=g(x)$, find $\dfrac{dy}{dx}$ by using $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$.
- 1 $y=3u-6$, $u=2x^2$
- 2 $y=6u^3$, $u=7x-4$
- 3 $y=\sin u$, $u=5x-1$
- 4 $y=\cos u$, $u=-\dfrac{x}{8}$
- 5 $y=\tan u$, $u=9x+2$
Two slopes multiplying into one
$u = g(x)$
$y = f(u)$
$y = f(g(x))$
$x$—
$u$—
du/dx—
dy/du—
product—
$dy/dx$ measured—
💡 Show solutions▶
1 $\dfrac{dy}{du}=3$ and $\dfrac{du}{dx}=4x$, so $\dfrac{dy}{dx}=12x$.
2 $\dfrac{dy}{du}=18u^2$ and $\dfrac{du}{dx}=7$, so
$$\frac{dy}{dx}=126u^2=126(7x-4)^2.$$
3 $\dfrac{dy}{du}=\cos u$ and $\dfrac{du}{dx}=5$, so $\dfrac{dy}{dx}=5\cos(5x-1)$.
4 $\dfrac{dy}{du}=-\sin u$ and $\dfrac{du}{dx}=-\dfrac18$, so
$$\frac{dy}{dx}=\frac18\sin\left(-\frac{x}{8}\right)=-\frac18\sin\left(\frac{x}{8}\right).$$
5 $\dfrac{dy}{du}=\sec^2u$ and $\dfrac{du}{dx}=9$, so $\dfrac{dy}{dx}=9\sec^2(9x+2)$.
Always substitute back The answer must be a function of $x$, so replace $u$ by $g(x)$ at the end.
II. Find $\dfrac{dy}{dx}$
II
- 1 $y=\ln(\sec x)$
- 2 $y=\csc(\pi x+1)$
- 3 $y=\cot^2x$
- 4 $y=\cos^3(\pi x)$
- 5 $y=\sqrt{\csc^{-1}x}$
- 6 $y=\dfrac{1}{\sin^2x}$
- 7 $y=\cos^{-1}\left(\sqrt{x}\right)$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
1 $\dfrac{dy}{dx}=\dfrac{1}{\sec x}\cdot\sec x\tan x=\tan x$.
2 $\dfrac{dy}{dx}=-\csc(\pi x+1)\cot(\pi x+1)\cdot\pi=-\pi\csc(\pi x+1)\cot(\pi x+1)$.
3 Write $y=(\cot x)^2$: $\dfrac{dy}{dx}=2\cot x\cdot\left(-\csc^2x\right)=-2\cot x\csc^2x$.
4 Write $y=[\cos(\pi x)]^3$: $\dfrac{dy}{dx}=3\cos^2(\pi x)\cdot(-\sin(\pi x))\cdot\pi=-3\pi\sin(\pi x)\cos^2(\pi x)$.
5 $y=\left(\csc^{-1}x\right)^{1/2}$, so
$$\frac{dy}{dx}=\frac{1}{2\sqrt{\csc^{-1}x}}\cdot\frac{-1}{|x|\sqrt{x^2-1}}=\frac{-1}{2|x|\sqrt{x^2-1}\sqrt{\csc^{-1}x}}.$$
6 Write $y=(\sin x)^{-2}$:
$$\frac{dy}{dx}=-2(\sin x)^{-3}\cos x=\frac{-2\cos x}{\sin^3x}.$$
7 $\dfrac{dy}{dx}=\dfrac{-1}{\sqrt{1-\left(\sqrt{x}\right)^2}}\cdot\dfrac{1}{2\sqrt{x}}=\dfrac{-1}{2\sqrt{x}\sqrt{1-x}}$.
II
Continued.
- 8 $y=\sin(\cos 7x)$
- 9 $y=\cot^{-1}\sqrt{4-x^2}$
- 10 $y=\cot^3(4x+1)$
- 11 $y=e^{\sin 3x}$
- 12 $y=(3x-2)^6$
- 13 $y=\left(3x^2+1\right)^3$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
8 Three layers — sine, cosine, then $7x$:
$$\frac{dy}{dx}=\cos(\cos 7x)\cdot(-\sin 7x)\cdot7=-7\sin(7x)\cos(\cos 7x).$$
9 With $g(x)=\sqrt{4-x^2}$ and $g'(x)=\dfrac{-x}{\sqrt{4-x^2}}$:
$$\frac{dy}{dx}=\frac{-1}{1+\left(4-x^2\right)}\cdot\frac{-x}{\sqrt{4-x^2}}=\frac{x}{\left(5-x^2\right)\sqrt{4-x^2}}.$$
10 $y=[\cot(4x+1)]^3$, so
$$\frac{dy}{dx}=3\cot^2(4x+1)\cdot\left(-\csc^2(4x+1)\right)\cdot4=-12\cot^2(4x+1)\csc^2(4x+1).$$
11 $\dfrac{dy}{dx}=e^{\sin3x}\cdot\cos(3x)\cdot3=3e^{\sin3x}\cos3x$.
12 $\dfrac{dy}{dx}=6(3x-2)^5\cdot3=18(3x-2)^5$.
13 $\dfrac{dy}{dx}=3\left(3x^2+1\right)^2\cdot6x=18x\left(3x^2+1\right)^2$.
III. Find the derivative
III
- 1 $y=\left(3x^2+3x-1\right)^4$
- 2 $y=(5-2x)^{-2}$
- 3 $f(x)=\sin^{-1}\left(x^2\right)$
- 4 $y=\left(2x^3-x^2+6x+1\right)^3$
- 5 $g(x)=(\tan x+\sin x)^{-3}$
- 6 $y=\left(\dfrac{x}{7}+\dfrac{7}{x}\right)^7$
- 7 $k(x)=\left(1+\tan^{-1}x\right)^3$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
1 $\dfrac{dy}{dx}=4\left(3x^2+3x-1\right)^3(6x+3)=12(2x+1)\left(3x^2+3x-1\right)^3$.
2 $\dfrac{dy}{dx}=-2(5-2x)^{-3}\cdot(-2)=\dfrac{4}{(5-2x)^3}$.
3 $f'(x)=\dfrac{1}{\sqrt{1-\left(x^2\right)^2}}\cdot2x=\dfrac{2x}{\sqrt{1-x^4}}$.
4 $\dfrac{dy}{dx}=3\left(2x^3-x^2+6x+1\right)^2\left(6x^2-2x+6\right)=6\left(3x^2-x+3\right)\left(2x^3-x^2+6x+1\right)^2$.
5 $g'(x)=-3(\tan x+\sin x)^{-4}\left(\sec^2x+\cos x\right)=\dfrac{-3\left(\sec^2x+\cos x\right)}{(\tan x+\sin x)^4}$.
6 $\dfrac{dy}{dx}=7\left(\dfrac{x}{7}+\dfrac{7}{x}\right)^6\left(\dfrac17-\dfrac{7}{x^2}\right)$.
7 $k'(x)=3\left(1+\tan^{-1}x\right)^2\cdot\dfrac{1}{1+x^2}$.
III
Continued.
- 8 $h(x)=x^2\cos^4x$
- 9 $y=\sin^5x$
- 10 $y=\tan(\sec x)$
- 11 $f(x)=\dfrac{e^{-x}}{x}$
- 12 $y=e^{2x+5}$
- 13 $g(x)=\sec^{-1}\left(\dfrac1x\right)$
- 14 $f(x)=\cot^{-1}\sqrt{4-x^2}$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
8 Product rule with the chain rule inside:
$$h'(x)=2x\cos^4x+x^2\cdot4\cos^3x(-\sin x)=2x\cos^3x\left(\cos x-2x\sin x\right).$$
9 $\dfrac{dy}{dx}=5\sin^4x\cos x$.
10 $\dfrac{dy}{dx}=\sec^2(\sec x)\cdot\sec x\tan x$.
11 Quotient rule with $\left(e^{-x}\right)'=-e^{-x}$:
$$f'(x)=\frac{-e^{-x}\cdot x-e^{-x}}{x^2}=\frac{-e^{-x}(x+1)}{x^2}.$$
12 $\dfrac{dy}{dx}=2e^{2x+5}$.
13 With $g'(u)=\dfrac{1}{|u|\sqrt{u^2-1}}$ at $u=\dfrac1x$ and $\dfrac{du}{dx}=-\dfrac{1}{x^2}$, the algebra collapses neatly:
$$g'(x)=\frac{1}{\left|\tfrac1x\right|\sqrt{\tfrac{1}{x^2}-1}}\cdot\left(-\frac{1}{x^2}\right)=\frac{-1}{\sqrt{1-x^2}}\quad(0\lt x\lt 1).$$
This is no accident: $\sec^{-1}\left(\tfrac1x\right)=\cos^{-1}x$.
14 Same as II.9: $f'(x)=\dfrac{x}{\left(5-x^2\right)\sqrt{4-x^2}}$.
III
Continued — the last seven.
- 15 $f(x)=\sqrt{e^{2x}+2x}$
- 16 $f(x)=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$
- 17 $g(x)=x^2\ln 9x$
- 18 $h(x)=\ln\left(4x^3+x\right)$
- 19 $g(x)=4xe^{x^2-1}$
- 20 $g(x)=\log_7\left(6x^4+3\right)^5$
- 21 $f(x)=2^x\cdot\log_3 7^{x^2-4}$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
15 $f(x)=\left(e^{2x}+2x\right)^{1/2}$, so
$$f'(x)=\frac{2e^{2x}+2}{2\sqrt{e^{2x}+2x}}=\frac{e^{2x}+1}{\sqrt{e^{2x}+2x}}.$$
16 Quotient rule, noting $\left(e^{-x}\right)'=-e^{-x}$:
$$f'(x)=\frac{\left(e^x+e^{-x}\right)^2-\left(e^x-e^{-x}\right)^2}{\left(e^x+e^{-x}\right)^2}=\frac{4}{\left(e^x+e^{-x}\right)^2}.$$
(This function is $\tanh x$, and the answer is $\operatorname{sech}^2x$.)
17 Product rule with $\dfrac{d}{dx}\ln9x=\dfrac1x$:
$$g'(x)=2x\ln 9x+x^2\cdot\frac1x=2x\ln9x+x=x\left(2\ln 9x+1\right).$$
18 $h'(x)=\dfrac{12x^2+1}{4x^3+x}$.
19 Product rule with the chain rule in the exponential:
$$g'(x)=4e^{x^2-1}+4x\cdot e^{x^2-1}(2x)=\left(8x^2+4\right)e^{x^2-1}.$$
20 Use the logarithm law first: $g(x)=5\log_7\left(6x^4+3\right)$, so
$$g'(x)=\frac{5}{\left(6x^4+3\right)\ln7}\cdot24x^3=\frac{120x^3}{\left(6x^4+3\right)\ln7}=\frac{40x^3}{\left(2x^4+1\right)\ln7}.$$
21 Simplify the logarithm first: $\log_37^{x^2-4}=\left(x^2-4\right)\dfrac{\ln7}{\ln3}$, so $f(x)=\dfrac{\ln7}{\ln3}\,2^x\left(x^2-4\right)$ and, by the product rule,
$$f'(x)=\frac{\ln7}{\ln3}\,2^x\left[\left(x^2-4\right)\ln2+2x\right].$$
IV–VI. More practice with the chain rule
IV
Find $\dfrac{dy}{dx}$. Parts 6–8 are compositions of three functions.
- 1 $y=\sqrt{5x^2+3}$
- 2 $y=\dfrac{4}{\sqrt[3]{2x-7}}$
- 3 $y=\sin^2(3x)$
- 4 $y=e^{\cos 2x}$
- 5 $y=\ln\left(x^2+4x+5\right)$
- 6 $y=\sqrt{\tan 4x}$
- 7 $y=\ln\left(\cos\left(x^2\right)\right)$
- 8 $y=e^{\sqrt{x^2+1}}$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
1 $y=\left(5x^2+3\right)^{1/2}$, so
$$\frac{dy}{dx}=\frac12\left(5x^2+3\right)^{-1/2}\cdot10x=\frac{5x}{\sqrt{5x^2+3}}.$$
2 $y=4(2x-7)^{-1/3}$, so
$$\frac{dy}{dx}=4\cdot\left(-\frac13\right)(2x-7)^{-4/3}\cdot2=-\frac{8}{3(2x-7)^{4/3}}.$$
3 $y=[\sin(3x)]^2$, so
$$\frac{dy}{dx}=2\sin(3x)\cdot\cos(3x)\cdot3=6\sin3x\cos3x=3\sin6x.$$
4 The exponent is the inner function: $\dfrac{dy}{dx}=e^{\cos2x}\cdot(-\sin2x)\cdot2=-2\sin(2x)\,e^{\cos2x}$.
5 $\dfrac{dy}{dx}=\dfrac{1}{x^2+4x+5}\cdot(2x+4)=\dfrac{2x+4}{x^2+4x+5}$. (The inside $x^2+4x+5=(x+2)^2+1$ is always positive, so the logarithm is defined for every $x$.)
6 Layers: square root $\circ$ tangent $\circ$ $4x$:
$$\frac{dy}{dx}=\frac{1}{2\sqrt{\tan4x}}\cdot\sec^2(4x)\cdot4=\frac{2\sec^2 4x}{\sqrt{\tan4x}}.$$
7 Layers: logarithm $\circ$ cosine $\circ$ $x^2$:
$$\frac{dy}{dx}=\frac{1}{\cos\left(x^2\right)}\cdot\left(-\sin\left(x^2\right)\right)\cdot2x=-2x\tan\left(x^2\right).$$
8 Layers: exponential $\circ$ square root $\circ$ $x^2+1$:
$$\frac{dy}{dx}=e^{\sqrt{x^2+1}}\cdot\frac{1}{2\sqrt{x^2+1}}\cdot2x=\frac{x\,e^{\sqrt{x^2+1}}}{\sqrt{x^2+1}}.$$
V
Combine the chain rule with the product or quotient rule. Simplify by factoring.
- 1 $y=x^3(2x+1)^4$
- 2 $y=\dfrac{x}{\sqrt{x^2+1}}$
- 3 $y=e^{-2x}\sin 3x$
- 4 $y=\dfrac{(x-1)^2}{(x+2)^3}$
- 5 $y=x\ln\left(x^2+1\right)$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
1 Product rule, with the chain rule on the second factor:
$$\frac{dy}{dx}=3x^2(2x+1)^4+x^3\cdot4(2x+1)^3\cdot2=x^2(2x+1)^3\left[3(2x+1)+8x\right]=x^2(2x+1)^3(14x+3).$$
2 Write $y=x\left(x^2+1\right)^{-1/2}$ and use the product rule:
$$\frac{dy}{dx}=\left(x^2+1\right)^{-1/2}+x\cdot\left(-\frac12\right)\left(x^2+1\right)^{-3/2}\cdot2x=\left(x^2+1\right)^{-3/2}\left[\left(x^2+1\right)-x^2\right]=\frac{1}{\left(x^2+1\right)^{3/2}}.$$
3 Product rule with $\left(e^{-2x}\right)'=-2e^{-2x}$ and $(\sin3x)'=3\cos3x$:
$$\frac{dy}{dx}=-2e^{-2x}\sin3x+3e^{-2x}\cos3x=e^{-2x}\left(3\cos3x-2\sin3x\right).$$
4 Quotient rule, then cancel the common factor $(x-1)(x+2)^2$:
$$\frac{dy}{dx}=\frac{2(x-1)(x+2)^3-(x-1)^2\cdot3(x+2)^2}{(x+2)^6}=\frac{(x-1)\left[2(x+2)-3(x-1)\right]}{(x+2)^4}=\frac{(x-1)(7-x)}{(x+2)^4}.$$
5 Product rule with $\dfrac{d}{dx}\ln\left(x^2+1\right)=\dfrac{2x}{x^2+1}$:
$$\frac{dy}{dx}=\ln\left(x^2+1\right)+x\cdot\frac{2x}{x^2+1}=\ln\left(x^2+1\right)+\frac{2x^2}{x^2+1}.$$
VI
Parts 1–5 use the table of values below.
| $x$ | $f(x)$ | $g(x)$ | $f'(x)$ | $g'(x)$ |
|---|---|---|---|---|
| $1$ | $3$ | $2$ | $-1$ | $4$ |
| $2$ | $1$ | $3$ | $5$ | $-2$ |
| $3$ | $2$ | $1$ | $7$ | $6$ |
- 1 $h(x)=f\big(g(x)\big)$. Find $h'(1)$.
- 2 $k(x)=g\big(f(x)\big)$. Find $k'(1)$.
- 3 $p(x)=\big(f(x)\big)^3$. Find $p'(2)$.
- 4 $r(x)=f\big(f(x)\big)$. Find $r'(2)$.
- 5 $s(x)=g\left(x^2\right)$. Find $s'(1)$.
- 6 Ripple. A stone dropped into a pond sends out a circular ripple whose radius grows as $r(t)=40t$ cm, $t$ in seconds. Find the rate at which the enclosed area $A=\pi r^2$ is increasing when $t=2$ s.
- 7 Compound interest. OMR 5000 is invested at $6\%$ per year compounded monthly, so the balance after $t$ years is $B(t)=5000\left(1+\dfrac{0.06}{12}\right)^{12t}$. Find $B'(t)$ and the rate at which the balance is growing after $5$ years.
- 8 Conceptual. A student writes $\dfrac{d}{dx}\left[\ln\left(x^2+1\right)\right]=\dfrac{1}{x^2+1}$ and $\dfrac{d}{dx}\big[f(2x)\big]=f'(2x)$. What is the common mistake, and what are the correct derivatives?
💡 Show solutions▶
1 $h'(1)=f'\big(g(1)\big)\cdot g'(1)=f'(2)\cdot g'(1)=5\cdot4=20$.
2 $k'(1)=g'\big(f(1)\big)\cdot f'(1)=g'(3)\cdot f'(1)=6\cdot(-1)=-6$.
3 Power rule for compositions: $p'(2)=3\big(f(2)\big)^2\cdot f'(2)=3(1)^2\cdot5=15$.
4 $r'(2)=f'\big(f(2)\big)\cdot f'(2)=f'(1)\cdot f'(2)=(-1)\cdot5=-5$.
5 The inner function is $x^2$, with derivative $2x$: $s'(x)=g'\left(x^2\right)\cdot2x$, so $s'(1)=g'(1)\cdot2=4\cdot2=8$.
6 By the chain rule, $\dfrac{dA}{dt}=\dfrac{dA}{dr}\cdot\dfrac{dr}{dt}=2\pi r\cdot40$. At $t=2$, $r=80$ cm, so
$$\frac{dA}{dt}=2\pi(80)(40)=6400\pi\approx20\,106\ \text{cm}^2/\text{s}.$$
(Check: $A(t)=1600\pi t^2$, so $A'(t)=3200\pi t$ and $A'(2)=6400\pi$.)
7 Write $B(t)=5000\,(1.005)^{12t}$. Using $\dfrac{d}{dt}\left[a^{u}\right]=a^{u}\ln a\cdot u'$ with $u=12t$:
$$B'(t)=5000\,(1.005)^{12t}\cdot\ln(1.005)\cdot12=60\,000\ln(1.005)\,(1.005)^{12t}.$$
At $t=5$: $B'(5)=60\,000\ln(1.005)(1.005)^{60}\approx403.65$ OMR per year (the balance itself is about OMR 6744.25).
8 Both forget to multiply by the derivative of the inner function. Correctly,
$$\frac{d}{dx}\left[\ln\left(x^2+1\right)\right]=\frac{2x}{x^2+1},\qquad \frac{d}{dx}\big[f(2x)\big]=2f'(2x).$$
🔬 Research–Teaching NexusResearch-oriented · optional, about 30 minutes
Research question. Does the chain rule $\frac{dF}{dt}=\frac{dF}{ds}\cdot\frac{ds}{dt}$ correctly predict fuel use per hour on a car trip?
Why it matters. Vehicle engineers and logistics companies combine linked rates like this to plan fuel, charging stops and delivery routes.
What to do.
- On a trip (e.g. Nizwa to Muscat, about 160 km) read the trip computer, or use sample data: fuel use 7.5 L per 100 km and average speed 110 km/h.
- Compute $\frac{dF}{ds}$ in L/km and $\frac{ds}{dt}$ in km/h.
- Use the chain rule to predict $\frac{dF}{dt}$ in L/h.
- Check with measurements: total fuel used divided by total driving time (sample: 12 L in 1.5 h, which is 8 L/h).
Tools: Calculator, car trip computer or sample data
Hand in: A half-page note: the predicted and measured rates, the units at each step, and one reason they differ.
Going further. Anywhere: repeat for walking, where calories per km times km per hour gives calories per hour.
Instructor note: Students should check units cancel (L/km $\times$ km/h $=$ L/h); the sample prediction is $0.075\times110=8.25$ L/h, close to 8 L/h.
🧪 Self-check quiz
0/5 correct
Q1. $\dfrac{d}{dx}\left[\sin\left(x^3\right)\right]$ is:
Q2. For $y=(3x-2)^6$, $\dfrac{dy}{dx}$ is:
Q3. If $y=f(u)$ and $u=g(x)$ with $\dfrac{dy}{du}=4$ and $\dfrac{du}{dx}=-3$ at a point, then $\dfrac{dy}{dx}$ there is:
Q4. $\dfrac{d}{dx}\left[e^{2x+5}\right]$ is:
Q5. Suppose $g(1)=2$, $g'(1)=4$, $f(2)=7$ and $f'(2)=-1$. If $h(x)=f\big(g(x)\big)$, then $h'(1)$ is: