Mc14, Mc15 - The chain rule, inverse derivatives and connected rates | OCR A-Level Mathematics B (MEI)

A rate may pass through several dependent quantities before reaching the one you want. Mc14 supplies the chain rule for composite functions; Mc15 uses it for inverse derivatives and connected rates. Learn to read the variable, sign and unit of each derivative before calculating.

How a change passes through a function

A circular oil film gains area at a steady $3\pi\,\mathrm{m^2\,min^{-1}}$ . Does its edge move outwards at a steady speed? At radius two metres the circumference is twice as large as at radius one, so the same new area is spread round twice as much edge. We will calculate the changing radius rate by linking derivatives.

You already know

Composite functions apply one function after another. You can differentiate powers, logarithms, exponentials and radian trigonometric functions. A product needs the product rule as well as any rule used inside its factors.

First take a numerical chain: $u=x^2+2$ , $y=u^4$ . At $x=1$ , a unit change in $x$ produces local rate two in $u$ ; each unit of $u$ then produces local rate $108$ in $y$ . Their combined local rate is $216$ .

Composite derivative

In $y=f(g(x))$ , the inner function $g$ supplies the input to the outer function $f$ . The derivative of the outer function must be evaluated at that supplied input.

This local-rate reasoning is made precise by differentiability: each function has a linear approximation with an error smaller than its input change. Composition multiplies the two linear coefficients.

Chain rule

If $u=g(x)$ is differentiable at $x$ , and $y=f(u)$ is differentiable at the corresponding $u$ ,

\frac{\mathrm dy}{\mathrm dx} =\frac{\mathrm dy}{\mathrm du}\frac{\mathrm du}{\mathrm dx}, \qquad [f(g(x))]'=f'(g(x))g'(x)

Not in the formula sheet: learn it.

Naming the intermediate variable makes both rates visible.

Worked example 1: a fourth power of a quadratic

Question. Differentiate $y=(x^2+2)^4$ .

Plan. Differentiate the fourth power with respect to its input, then multiply by that input's rate.

\begin{aligned} u&=x^2+2,\quad \frac{\mathrm du}{\mathrm dx}=2x &&\text{inner rate}\\ \frac{\mathrm dy}{\mathrm du}&=4u^3 &&\text{outer rate}\\ \frac{\mathrm dy}{\mathrm dx}&=4(x^2+2)^3(2x)=8x(x^2+2)^3 &&\text{multiply} \end{aligned}

Why this step? The outer derivative is $4u^3$ , not $4x^3$ ; its current input is the whole quadratic.

Answer. $y'=8x(x^2+2)^3$ .

Check. At one, this is $8(27)=216$ , agreeing with the two local rates.

Change only the sign of the constant inside the quadratic.

Worked example 2: changing the inner shift

Question. Differentiate $y=(x^2-2)^4$ .

Plan. Keep the inner derivative $2x$ , but evaluate the outer derivative at the new input.

\begin{aligned} u&=x^2-2,\quad u'=2x &&\text{inner derivative}\\ y'&=4u^3(2x) &&\text{chain rule}\\ &=8x(x^2-2)^3 &&\text{substitute} \end{aligned}

Why this step? A constant shift does not alter the inner rate, but it changes the value entering the outer derivative.

Answer. $y'=8x(x^2-2)^3$ .

Check. At one the derivative is minus eight; at zero it is zero because the inner rate is zero.

The notation resembles cancelling $\mathrm du$ , but the rule is a theorem about rates, not ordinary fraction cancellation. In particular it remains valid at a zero inner derivative. The earlier cubic example $(x^2+1)^3$ gives $6x(x^2+1)^2$ , illustrating the same structure.

Watch out

Wrong: keep only the outer derivative $4(x^2-2)^3$ . That is a rate with respect to $u$ ; multiply by $2x$ to obtain the rate with respect to $x$ .

Change just the inner constant again.

Choosing rules for composite expressions

Look at the outside operation before choosing a derivative. A logarithm, exponential or sine may all have a quadratic input, yet their outer derivatives are different.

Worked example 3: a logarithm of a quadratic

Question. Differentiate $f(x)=\ln(4-x^2)$ , giving its real domain.

Plan. Require a positive logarithm input, then multiply its reciprocal by the inner derivative.

\begin{aligned} 4-x^2&>0\iff -2<x<2 &&\text{domain}\\ f'(x)&=\frac1{4-x^2}(-2x) &&\text{chain rule}\\ &=-\frac{2x}{4-x^2} &&\text{simplify} \end{aligned}

Why this step? The logarithm requires positivity, not just the nonzero denominator suggested by the final fraction.

Answer. $f'(x)=-2x/(4-x^2)$ , for $-2<x<2$ .

Check. At zero the derivative is zero; on $0<x<2$ , the shrinking logarithm input gives a negative derivative.

Change only the sign of the quadratic inside the logarithm.

Worked example 4: a positive input everywhere

Question. Differentiate $g(x)=\ln(4+x^2)$ , giving its real domain.

Plan. Check positivity, then use the same reciprocal-input outer derivative.

\begin{aligned} 4+x^2&>0 &&\text{every real input}\\ g'(x)&=\frac1{4+x^2}(2x) &&\text{chain rule}\\ &=\frac{2x}{4+x^2} &&\text{simplify} \end{aligned}

Why this step? The new input is never zero, so neither the function nor its derivative has a missing real input.

Answer. $g'(x)=2x/(4+x^2)$ , for all real $x$ .

Check. It is positive for $x>0$ , matching the increasing input.

For a root, $(\sqrt{5-x^2})'=-x/\sqrt{5-x^2}$ : the function includes its endpoints $\pm\sqrt5$ , but its finite derivative does not. Other outer functions give $(\mathrm e^{-x^2})'=-2x\mathrm e^{-x^2}$ and $(\sin(x^2))'=2x\cos(x^2)$ , using radians. A product around a composite needs both rules:

(x\mathrm e^{-x^2})'=\mathrm e^{-x^2}+x(-2x\mathrm e^{-x^2}) =\mathrm e^{-x^2}(1-2x^2)

The zeros are $\pm1/\sqrt2$ , because the exponential never vanishes. Replacing four by three in the first logarithm gives domain $(-\sqrt3,\sqrt3)$ and derivative $-2x/(3-x^2)$ .

Problem-solving

Read from outside to inside to select the rules; calculate from inside to outside to assemble the derivative. Keep function-domain restrictions alongside the result.

Alter the logarithm's constant but keep its negative quadratic.

Reversing a derivative on a valid branch

You may be given $x$ in terms of $y$ even though you need the rate of $y$ with respect to $x$ . A valid local inverse and a nonzero forward derivative let you reverse the rate. Differentiating $h(y(x))=x$ gives $h'(y)y'=1$ .

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Mc14, Mc15 - The chain rule, inverse derivatives and connected rates

How a change passes through a function

Choosing rules for composite expressions

Reversing a derivative on a valid branch

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