1.07r - The chain rule, inverse derivatives and connected rates

1.07r - The chain rule, inverse derivatives and connected rates

One changing quantity can drive another through several stages. Use 1.07r to differentiate composite functions, reverse a derivative on a valid inverse branch and connect instantaneous rates. The variable, sign and unit of each rate matter as much as the arithmetic.

How a change passes through a function

If air enters a spherical balloon at a constant volume rate, does its radius grow at a constant rate too? At radius three centimetres, an extra cubic centimetre produces much less radial change than it would in a tiny balloon. We need a rule that passes a rate through the volume–radius relationship.

You already know

A composite function applies one function after another. Differentiate powers, exponentials, logarithms and radian trigonometric functions; use the product rule when two factors multiply.

Begin with u=x2+1u=x^2+1 and y=u3y=u^3. At x=1x=1, the inner rate is two and the outer rate is twelve, so the composite rate is twenty-four. Each unit of input change is amplified twice.

Inner and outer functions

For y=f(g(x))y=f(g(x)), gg is the inner function and ff is the outer function. In (x2+1)3(x^2+1)^3, the outer function cubes its input; it is not the function x3x^3 applied directly to the initial xx.

Multiplying local rates gives the formal rule, provided the inner function is differentiable at the input and the outer function is differentiable at its resulting input.

Chain rule

If u=g(x)u=g(x) and y=f(u)y=f(u), then

dydx=dydududx=f′(g(x))g′(x)\frac{\mathrm dy}{\mathrm dx} =\frac{\mathrm dy}{\mathrm du}\frac{\mathrm du}{\mathrm dx} =f'(g(x))g'(x)

Not in the formula sheet: learn it.

The outer derivative must be evaluated at the inner value.

Worked example 1: two stages of a power

Question. Differentiate y=(x2+1)3y=(x^2+1)^3.

Plan. Name the inner expression uu, then multiply the two derivatives.

u=x2+1,dudx=2xinner ratedydu=3u2outer ratedydx=3(x2+1)2(2x)=6x(x2+1)2multiply\begin{aligned} u&=x^2+1,\quad \frac{\mathrm du}{\mathrm dx}=2x &&\text{inner rate}\\ \frac{\mathrm dy}{\mathrm du}&=3u^2 &&\text{outer rate}\\ \frac{\mathrm dy}{\mathrm dx}&=3(x^2+1)^2(2x)=6x(x^2+1)^2 &&\text{multiply} \end{aligned}

Why this step? The outer function sees x2+1x^2+1, so that whole expression stays inside its derivative.

Answer. y′=6x(x2+1)2y'=6x(x^2+1)^2.

Check. At zero the inner rate is zero, so the composite rate is zero too; at one the derivative is twenty-four.

Change only the coefficient of x2x^2 in the inner function, from one to two.

Worked example 2: a faster inner change

Question. Differentiate y=(2x2+1)3y=(2x^2+1)^3.

Plan. Update the inner rate while keeping the outer power rule.

u=2x2+1,u′=4xinner derivativey′=3u2(4x)chain rule=12x(2x2+1)2return to x\begin{aligned} u&=2x^2+1,\quad u'=4x &&\text{inner derivative}\\ y'&=3u^2(4x) &&\text{chain rule}\\ &=12x(2x^2+1)^2 &&\text{return to }x \end{aligned}

Why this step? The inner derivative is 4x4x; retaining just the outer derivative would measure change with respect to uu, not xx.

Answer. y′=12x(2x2+1)2y'=12x(2x^2+1)^2.

Check. At one this gives 108108; expanding the cube before differentiating gives the same value.

The differential notation looks like cancellation, but the rule is valid even when the inner derivative is zero. More precisely, differentiability makes each function locally linear plus a smaller error; composing those approximations multiplies their linear coefficients.

Watch out

Wrong: ((x2+1)3)′=3x2((x^2+1)^3)'=3x^2. Right: 3(x2+1)2(2x)3(x^2+1)^2(2x). Keep the whole inner expression and multiply by its derivative.

Change that inner coefficient again.

Choosing rules for composite expressions

Identify the outer operation first: square root, exponential, logarithm or sine. Then check where both the function and the differentiated expression are real and finite.

Worked example 3: a square root with an inner quadratic

Question. Differentiate y=5−x2y=\sqrt{5-x^2}, giving the function and finite-derivative domains.

Plan. Write the root as a power and multiply by the inner derivative.

5−x2≥0  ⟺  −5≤x≤5function domainy′=12(5−x2)−1/2(−2x)chain rule=−x5−x2simplify\begin{aligned} 5-x^2&\ge0\iff -\sqrt5\le x\le\sqrt5 &&\text{function domain}\\ y'&=\tfrac12(5-x^2)^{-1/2}(-2x) &&\text{chain rule}\\ &=-\frac{x}{\sqrt{5-x^2}} &&\text{simplify} \end{aligned}

Why this step? The root exists at its zero endpoints, but the reciprocal root in the derivative does not.

Answer. The function exists on the closed interval; the finite derivative exists on −5<x<5-\sqrt5<x<\sqrt5.

Check. y′(0)=0y'(0)=0, fitting the top of the upper semicircle; its endpoints have vertical tangents.

Change only the sign of the quadratic inside the root. That changes the domain dramatically.

Worked example 4: a root with no zero input

Question. Repeat for y=5+x2y=\sqrt{5+x^2}.

Plan. Differentiate the power and check the radicand.

5+x2>0all real xy′=12(5+x2)−1/2(2x)chain rule=x5+x2simplify\begin{aligned} 5+x^2&>0 &&\text{all real }x\\ y'&=\tfrac12(5+x^2)^{-1/2}(2x) &&\text{chain rule}\\ &=\frac{x}{\sqrt{5+x^2}} &&\text{simplify} \end{aligned}

Why this step? A strictly positive radicand keeps both the function and its reciprocal-root derivative defined everywhere.

Answer. y′=x/5+x2y'=x/\sqrt{5+x^2}, for all real xx.

Check. The derivative is zero at zero and positive for positive inputs, matching the rising right-hand branch.

The same rule gives (e−x2)′=−2xe−x2(\mathrm e^{-x^2})'=-2x\mathrm e^{-x^2}, (ln⁡(3−x2))′=−2x/(3−x2)(\ln(3-x^2))'=-2x/(3-x^2) on −3<x<3-\sqrt3<x<\sqrt3, and (sin⁡(x2))′=2xcos⁡(x2)(\sin(x^2))'=2x\cos(x^2) in radians. For y=xe−x2y=x\mathrm e^{-x^2}, use the product rule outside and the chain rule inside:

y′=e−x2+x(−2xe−x2)=e−x2(1−2x2)y'=\mathrm e^{-x^2}+x(-2x\mathrm e^{-x^2}) =\mathrm e^{-x^2}(1-2x^2)

Its derivative vanishes only at x=±1/2x=\pm1/\sqrt2, since the exponential is positive.

Watch out

A logarithm needs a positive input, not merely a nonzero denominator in its derivative. A square root may have a larger function domain than derivative domain.

Return to the root and alter only its constant.

Reversing a derivative on a valid branch

Sometimes the easy direction is xx in terms of yy. If it has a differentiable local inverse and a nonzero derivative, differentiate x=h(y(x))x=h(y(x)) to obtain h′(y)y′=1h'(y)y'=1.

Continue this lesson.

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