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)$^5$”, the inner function 内函数 is $2x^2+1$:

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

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$1$	$4$	$-2$	$3$	$5$
$2$	$1$	$6$	$-1$	$2$
$3$	$7$	$3$	$2$	$-4$

For $k(x)=f\big(g(x)\big)$, find the tangent line at $x=3$. The point is $k(3)=f\big(g(3)\big)=f(2)=1$. The slope is $k'(3)=f'\big(g(3)\big)\,g'(3)=f'(2)(-4)=6(-4)=-24$. So the tangent is $y-1=-24(x-3)$.

For $p(x)=f(x^2-x)$ at $x=2$: the inside is $x^2-x=2$ and its derivative is $2x-1=3$, so $p'(2)=f'(2)\cdot3=18$. For $h(x)=\ln\big(f(x)\big)$: $h'(x)=\dfrac{f'(x)}{f(x)}$, so $h'(3)=\dfrac{3}{7}$. For a product with a chain inside, $m(x)=f(2x)\,g(x)$, use both rules: $m'(x)=2f'(2x)\,g(x)+f(2x)\,g'(x)$, so $m'(1)=2f'(2)\,g(1)+f(2)\,g'(1)=2(6)(3)+1(5)=41$. The inner derivative $2$ from $f(2x)$ is the factor most often lost.

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 line 切线 to the circle $x^2+y^2=25$ at the point $(3,4)$. From above $\dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac{3}{4}$ there, so the tangent is

Notice the tangent is perpendicular 垂直 to the radius 半径, as geometry promises – a good sanity check.

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). For the curve $y^3-xy=2$:

At $(-1,1)$ the slope is $\dfrac{1}{3+1}=\dfrac14$, so the tangent is $y-1=\tfrac14(x+1)$.

• Horizontal tangent: numerator $y=0$. But $y=0$ gives $0-0=2$, which is false, so no point on the curve has a horizontal tangent.
• Vertical tangent: denominator $3y^2-x=0$, so $x=3y^2$. Substitute into the curve: $y^3-3y^3=2$, so $y^3=-1$, $y=-1$ and $x=3$. The tangent is vertical at $(3,-1)$; check the numerator there is not also $0$ ($y=-1\ne0$).

Always substitute back into the original equation: a point that makes the derivative zero or undefined is useless unless it is on the curve.

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 $g'(b)=\dfrac{1}{f'(a)}$.

You do not need a formula for the inverse. For $f(x)=x^3+x$, let $g=f^{-1}$; to find $g'(2)$, first solve $f(a)=2$: $a^3+a=2$ gives $a=1$. Then $g'(2)=\dfrac{1}{f'(1)}=\dfrac{1}{3(1)^2+1}=\dfrac14$.

Worked example. Suppose $f(2)=5$ and $f'(2)=3$, and $g$ is the inverse of $f$. Because $(2,5)$ lies on $f$, the point $(5,2)$ lies on $g$, and

The graphs of $f$ and $g$ are mirror images across $y=x$, so their slopes at matching points are reciprocals.

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. $\dfrac{d}{dx}\arctan(3x)=\dfrac{3}{1+9x^2}$. The co-functions are the negatives: $\dfrac{d}{dx}\arccos x=-\dfrac{1}{\sqrt{1-x^2}}$, and similarly for $\text{arccot}$ and $\text{arccsc}$.

Worked example. $\dfrac{d}{dx}\arcsin(x^2)=\dfrac{1}{\sqrt{1-(x^2)^2}}\cdot2x=\dfrac{2x}{\sqrt{1-x^4}}$. Where these formulas come from: if $y=\arcsin x$, then $\sin y=x$; differentiating implicitly, $\cos y\,\dfrac{dy}{dx}=1$, so $\dfrac{dy}{dx}=\dfrac{1}{\cos y}=\dfrac{1}{\sqrt{1-x^2}}$ (using $\cos y=\sqrt{1-\sin^2y}$, positive for $-\tfrac{\pi}{2}<y<\tfrac{\pi}{2}$).

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.

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 (second derivative at a point). For $y^3-xy=2$ at $(-1,1)$ we found $\dfrac{dy}{dx}=\dfrac14$. Differentiate $\dfrac{dy}{dx}=\dfrac{y}{3y^2-x}$ with the quotient rule:

Substitute numbers before simplifying: $\dfrac{\tfrac14(4)-1\left(6\cdot\tfrac14-1\right)}{4^2}=\dfrac{1-\tfrac12}{16}=\dfrac{1}{32}$.

From a differential equation. If $\dfrac{dy}{dx}=2x-y$, then $\dfrac{d^2y}{dx^2}=2-\dfrac{dy}{dx}=2-(2x-y)=2-2x+y$. In Quadrant II ($x<0$, $y>0$), $2-2x+y>0$, so every solution curve is concave up there.

Practice.

1. Differentiate $y=\cos^3(2x)$.
2. Use the table above: find $\dfrac{d}{dx}\,g\big(f(x)\big)$ at $x=2$.
3. For $x^2y+y^3=10$, find $\dfrac{dy}{dx}$ at $(3,1)$.
4. $f(x)=e^x+2x$ and $g=f^{-1}$. Find $g'(1)$.
5. Find $\dfrac{d}{dx}\arctan\left(\sqrt x\right)$.
6. Find $\dfrac{d^2y}{dx^2}$ for $y=x\ln x$.

Answers.

1. $3\cos^2(2x)\cdot(-\sin 2x)\cdot2=-6\cos^2(2x)\sin(2x)$.
2. $g'\big(f(2)\big)\,f'(2)=g'(1)\cdot6=5\cdot6=30$.
3. $2xy+x^2\dfrac{dy}{dx}+3y^2\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+3y^2}=-\dfrac{6}{12}=-\dfrac12$.
4. $f(0)=1$, so $g(1)=0$ and $g'(1)=\dfrac{1}{f'(0)}=\dfrac{1}{e^0+2}=\dfrac13$.
5. $\dfrac{1}{1+x}\cdot\dfrac{1}{2\sqrt x}=\dfrac{1}{2\sqrt x\,(1+x)}$.
6. $y'=\ln x+1$, so $y''=\dfrac1x$.

• 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.
• Horizontal tangent: numerator of $\tfrac{dy}{dx}$ is $0$; vertical tangent: denominator is $0$. Then check the point is on the curve.
• For an implicit second derivative at a point, substitute the numbers (including $\tfrac{dy}{dx}$) before simplifying.