Differentiation: Composite, Implicit, and Inverse Functions

Unit 2 differentiated single functions. Unit 3 differentiates functions built inside other functions. The chain rule 链式法则 differentiates a composite function 复合函数 $f\big(g(x)\big)$:

“Derivative of the outer function (leaving the inside alone), times the derivative of the inside.” The inner derivative $g'(x)$ is the piece students forget, so always ask “what is the inside, and what is its derivative?” Example:

In Leibniz notation 莱布尼茨记号, with $y=f(u)$ and $u=g(x)$, the rule reads $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$ – the intermediate $du$ appears to “cancel.” Exam questions often give a table for $f$, $g$, $f'$, $g'$ and ask for $h'(a)$ where $h(x)=f\big(g(x)\big)$; evaluate $f'\big(g(a)\big)\cdot g'(a)$ by reading values.

Worked example. Differentiate $h(x)=(2x^2+1)^5$. The outer function 外函数 is “(something)$^5$” and the inner function 内函数 is $2x^2+1$:

Worked example (from a table). The table gives values of $f$, $f'$, $g$ and $g'$.

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$-1$	$3$	$4$	$2$	$-3$
$2$	$-1$	$5$	$0$	$1$
$3$	$2$	$-2$	$-1$	$6$

(a) $p(x)=f(x^2-x)$. Find the slope of the tangent line at $x=-1$. The inside is $x^2-x=(-1)^2-(-1)=2$, and its derivative is $2x-1=-3$. So $p'(-1)=f'(2)\cdot(-3)=5(-3)=-15$.

(b) $k(x)=g\big(f(x)\big)$. Find $k'(3)$. The inside value is $f(3)=2$, so $k'(3)=g'\big(f(3)\big)\,f'(3)=g'(2)\cdot(-2)=1(-2)=-2$.

(c) $m(x)=\big(g(x)\big)^3$. Find $m'(-1)$. The outer function is "cube": $m'(x)=3\big(g(x)\big)^2g'(x)$, so $m'(-1)=3(2)^2(-3)=-36$.

Each answer reads two rows of the table: the row of the input $x$, and the row of the inside value. Students who use only one row get the wrong number.

Some curves are defined implicitly – by an equation in $x$ and $y$ that is not solved for $y$, such as $x^2+y^2=25$. Implicit differentiation 隐函数求导 finds $\dfrac{dy}{dx}$ without solving for $y$ first. It is just the chain rule, treating $y$ as a function of $x$.

The method: differentiate both sides with respect to $x$; every time you differentiate a $y$-term, multiply by $\dfrac{dy}{dx}$ (the chain rule); then solve algebraically for $\dfrac{dy}{dx}$. For $x^2+y^2=25$:

Worked example. Find the tangent to $x^2+y^2=25$ at $(3,4)$. Here $\dfrac{dy}{dx}=-\dfrac{3}{4}$, so the tangent line 切线 is $y-4=-\tfrac{3}{4}(x-3)$ – perpendicular 垂直 to the radius 半径, as geometry predicts.

Exam skill – “Show that $\dfrac{dy}{dx}=\ldots$”. This exact prompt appears most years (e.g. “Show that $\dfrac{dy}{dx}=\dfrac{2y}{y^2-2x}$”). Because the target is given, you must show every algebra step cleanly: differentiate both sides, use the product/chain rules on mixed $xy$ terms, collect all $\dfrac{dy}{dx}$ terms on one side, factor, and divide. A correct final line that skips the algebra earns little. Follow-up parts then ask for a tangent line, or where the tangent is horizontal ($\tfrac{dy}{dx}=0$, so the numerator is $0$) or vertical (the denominator is $0$).

Worked example. The curve $x^2+xy+y^2=7$ passes through $(1,2)$. Show that $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$, and find the slope at $(1,2)$.

The middle term $xy$ is a product, so it needs the product rule: $\dfrac{d}{dx}(xy)=1\cdot y+x\dfrac{dy}{dx}$. Differentiate every term:

Collect the $\dfrac{dy}{dx}$ terms on one side and factor:

At $(1,2)$ (check: $1+2+4=7$), the slope is $-\dfrac{2(1)+2}{1+2(2)}=-\dfrac{4}{5}$.

Common mistakes. Writing $\dfrac{d}{dx}(xy)=\dfrac{dy}{dx}$ (the product rule is missing), and writing $\dfrac{d}{dx}(y^2)=2y$ (the factor $\dfrac{dy}{dx}$ is missing).

Worked example (horizontal and vertical tangents). On the same curve $x^2+xy+y^2=7$, where is the tangent horizontal, and where is it vertical?

• Horizontal: the numerator is zero, $2x+y=0$, so $y=-2x$. Substitute into the original equation: $x^2-2x^2+4x^2=7$, so $3x^2=7$ and $x=\pm\sqrt{7/3}$. The points are $\left(\sqrt{7/3},\,-2\sqrt{7/3}\right)$ and $\left(-\sqrt{7/3},\,2\sqrt{7/3}\right)$.
• Vertical: the denominator is zero, $x+2y=0$, so $x=-2y$. Substitute: $4y^2-2y^2+y^2=7$, so $y=\pm\sqrt{7/3}$, at $\left(-2\sqrt{7/3},\,\sqrt{7/3}\right)$ and $\left(2\sqrt{7/3},\,-\sqrt{7/3}\right)$.

At each point check that the other part of the fraction is not also zero; $\tfrac00$ needs more work.

If $g$ is the inverse function 反函数 of $f$ (so $f(g(x))=x$), the chain rule links their derivatives:

In words: the derivative of the inverse at a point is the reciprocal 倒数 of the derivative of the original function at the matching point. A common exam setup gives a table and a point $(a,b)$ on $f$ (so $(b,a)$ is on $g$), then asks for $g'(b)=\dfrac{1}{f'(a)}$.

Worked example. If $f(2)=5$ and $f'(2)=3$, and $g$ is the inverse of $f$, then $(5,2)$ lies on $g$ and $g'(5)=\dfrac{1}{f'(2)}=\dfrac{1}{3}$.

Worked example (no formula for the inverse). $f(x)=x^5+2x+1$ and $g=f^{-1}$. Find $g'(4)$.

You cannot solve $y=x^5+2x+1$ for $x$, and you do not need to. First find the matching point: $f(a)=4$ means $a^5+2a+1=4$, and $a=1$ works. So $g(4)=1$, and

Why the formula is true. Differentiate $f\big(g(x)\big)=x$ with the chain rule: $f'\big(g(x)\big)\,g'(x)=1$. Divide by $f'\big(g(x)\big)$.

Common mistake. Writing $g'(4)=\dfrac{1}{f'(4)}$. The derivative of $f$ must be taken at the matching $x$-value $g(4)=1$, not at $4$.

The same idea gives the derivatives of the inverse trigonometric functions 反三角函数. The three you should know:

Combine these with the chain rule when the input is itself a function, e.g. $\dfrac{d}{dx}\arctan(3x)=\dfrac{3}{1+9x^2}$.

Where the formulas come from. Let $y=\arctan x$, so $\tan y=x$ with $-\tfrac{\pi}{2}<y<\tfrac{\pi}{2}$. Differentiate implicitly: $\sec^2y\,\dfrac{dy}{dx}=1$, so

The same steps with $\sin y=x$ give $\dfrac{1}{\cos y}=\dfrac{1}{\sqrt{1-x^2}}$. The co-functions are the negatives: $\dfrac{d}{dx}\arccos x=-\dfrac{1}{\sqrt{1-x^2}}$.

Worked example. $\dfrac{d}{dx}\Big(x\arcsin x\Big)=\arcsin x+\dfrac{x}{\sqrt{1-x^2}}$ (product rule). And $\dfrac{d}{dx}\arctan\left(e^{x}\right)=\dfrac{e^{x}}{1+e^{2x}}$ (chain rule: the inside is $e^x$).

A skill topic: real derivatives mix several rules, so read the structure of the expression first, from the outside in.

• Is it a sum? Differentiate term by term.
• A product or quotient? Apply the product rule 乘积法则 or the quotient rule 商法则, and expect to use the chain rule inside.
• A composite (something inside something)? Chain rule.
• Given implicitly? Implicit differentiation.

Name the outermost operation, apply its rule, and recurse inward. Neatness prevents the sign and bookkeeping errors that cost marks.

Worked example. Differentiate $y=\dfrac{\ln(x^2+1)}{x}$. Outermost: a quotient. Inside the numerator: a chain ($\ln$ of $x^2+1$).

Sometimes a rewrite saves a rule: $\dfrac{5}{x^3}=5x^{-3}$ needs only the power rule, $-15x^{-4}$, not the quotient rule.

Differentiating $f'$ produces the second derivative 二阶导数 $f''$; repeating gives higher-order derivatives 高阶导数. The notations:

The second derivative measures how the slope is changing; it drives concavity 凹凸性 and acceleration 加速度 in later units. To find $f''$ implicitly, differentiate the expression for $\dfrac{dy}{dx}$ again (with the quotient and chain rules), then substitute $\dfrac{dy}{dx}$ back in.

Worked example. For the circle $x^2+y^2=25$, find $\dfrac{d^2y}{dx^2}$ in terms of $y$.

Start from $\dfrac{dy}{dx}=-\dfrac{x}{y}$ and use the quotient rule, remembering that $y$ depends on $x$:

The last step uses the curve itself: $x^2+y^2=25$. On the top half ($y>0$) the second derivative is negative, so the curve is concave down there, as a circle's top should be.

Worked example (implicit, in terms of $x$ and $y$). For $x^2+xy+y^2=7$ we found $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$. The quotient rule gives

Substitute $\dfrac{dy}{dx}$: $y-x\dfrac{dy}{dx}=\dfrac{y(x+2y)+x(2x+y)}{x+2y}=\dfrac{2(x^2+xy+y^2)}{x+2y}=\dfrac{14}{x+2y}$, using the curve. So

Reading $f''$ from the graph of $f'$. $f''$ is the slope of the graph of $f'$. Where that graph has a corner, the left and right slopes differ, so $f''$ does not exist there.

A pattern of derivatives. For $f(x)=\ln(1+x)$:

Each step brings down the power as a factor and the sign alternates, so $f^{(n)}(x)=(-1)^{n+1}\dfrac{(n-1)!}{(1+x)^n}$. Patterns like this build Taylor polynomials in Unit 10: $f^{(4)}(0)=-6$.

• Use the chain rule for composite functions — differentiate the outside, then multiply by the derivative of the inside (the most-forgotten factor).
• For implicit differentiation, differentiate both sides with respect to $x$ and attach $\tfrac{dy}{dx}$ each time $y$ is differentiated, then solve.
• Get the second derivative by differentiating twice (velocity → acceleration).
• Combine rules carefully in layered expressions (chain inside product, etc.).
• The inverse function's graph is the reflection in $y=x$; its slope is the reciprocal of the original's at the matching point.
• For an inverse, find the matching point first: $g'(b)=\dfrac{1}{f'(a)}$ where $f(a)=b$.

Practice.

1. Differentiate $y=\sin^3(4x)$.
2. Use the table in the chain-rule section: find $\dfrac{d}{dx}f\big(g(x)\big)$ at $x=3$.
3. For $y^2-x^3=xy+1$, find $\dfrac{dy}{dx}$ at $(0,1)$.
4. $f(x)=x^3+4x-2$ and $g=f^{-1}$. Find $g'(3)$.
5. Find $\dfrac{d}{dx}\arcsin(3x)$ and $\dfrac{d}{dx}\arctan\!\left(\dfrac1x\right)$.
6. For $f(x)=e^{-2x}$, find $f^{(4)}(x)$ and a formula for $f^{(n)}(x)$.

Answers.

1. $3\sin^2(4x)\cdot\cos(4x)\cdot4=12\sin^2(4x)\cos(4x)$.
2. $f'\big(g(3)\big)\,g'(3)=f'(-1)\cdot6=4\cdot6=24$.
3. $2y\dfrac{dy}{dx}-3x^2=y+x\dfrac{dy}{dx}$, so $\dfrac{dy}{dx}=\dfrac{3x^2+y}{2y-x}=\dfrac{0+1}{2-0}=\dfrac12$.
4. $f(1)=3$, so $g(3)=1$ and $g'(3)=\dfrac{1}{f'(1)}=\dfrac{1}{3+4}=\dfrac17$.
5. $\dfrac{3}{\sqrt{1-9x^2}}$; $\dfrac{1}{1+x^{-2}}\cdot\left(-\dfrac{1}{x^2}\right)=-\dfrac{1}{x^2+1}$.
6. $f^{(4)}(x)=16e^{-2x}$; $f^{(n)}(x)=(-2)^ne^{-2x}$.