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, 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 is "(something)$^5$", the inner is $2x^2+1$:

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 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 to the radius, as geometry promises – a good sanity check.

Implicit differentiation gives the tangent to a circle, perpendicular to the radius

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 $0$).

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

The inverse function is the mirror image of the function in the line y = x

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. $\dfrac{d}{dx}\arctan(3x)=\dfrac{3}{1+9x^2}$.

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 that rule, 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. The notations:

The second derivative measures how the slope is changing; it drives concavity 凹凸性 and acceleration 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.

• 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.