P3 4.2 - Product and chain rules

P3 4.2 - Product and chain rules

Differentiate functions built by putting one function inside another or multiplying two functions together. Use their structure to choose the chain rule, the product rule or both, and check that no changing factor has been overlooked.

The chain rule

Can you differentiate (x2+1)20(x^2+1)^{20} without expanding it? The expression has an inside, x2+1x^2+1, and an outside, raising to the twentieth power. Keeping that structure saves a great deal of algebra.

Let u=x2+1u=x^2+1 and y=u20y=u^{20}. Near a particular point, a small change in xx produces approximately 2x2x times that change in uu. A small change in uu then produces approximately 20u1920u^{19} times that change in yy. Multiply those two local rates:

dydx=20u19×2x=40x(x2+1)19.\frac{dy}{dx}=20u^{19}\times2x=40x(x^2+1)^{19}.

Chain rule

If y=F(u)y=F(u) and u=g(x)u=g(x) are differentiable at the relevant points, then

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

Differentiate the outer function, keep its input unchanged, then multiply by the inner derivative. Not in the formula booklet: learn it.

The notation resembles cancelling dudu, but the reason for the rule is composition of rates. There can be several layers. With u=g(x)u=g(x), useful forms include

ddx(un)=nun−1u′,ddx(eu)=euu′,ddx(ln⁡u)=u′u,ddx(cos⁡u)=−sin⁡u u′.\begin{aligned} \frac{d}{dx}(u^n)&=nu^{n-1}u',& \frac{d}{dx}(e^u)&=e^u u',\\ \frac{d}{dx}(\ln u)&=\frac{u'}u,& \frac{d}{dx}(\cos u)&=-\sin u\,u'. \end{aligned}

Use each formula only where the original function is defined and differentiable; in particular ln⁡u\ln u requires u>0u>0.

Worked example 1: a positive outer power

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

Plan. Differentiate the cube first, then the quadratic inside.

u=x2+1,dudx=2x,y=u3,dydu=3u2,dydx=3(x2+1)2×2x=6x(x2+1)2.\begin{aligned} u&=x^2+1,&\frac{du}{dx}&=2x,\\ y&=u^3,&\frac{dy}{du}&=3u^2,\\ \frac{dy}{dx}&=3(x^2+1)^2\times2x=6x(x^2+1)^2. \end{aligned}

Answer. 6x(x2+1)26x(x^2+1)^2.

Check. Expanding first gives y=x6+3x4+3x2+1y=x^6+3x^4+3x^2+1, whose derivative 6x5+12x3+6x6x^5+12x^3+6x factors to the same result.

Change only the outer exponent from 33 to −3-3. The chain rule still applies without expansion.

Worked example 2: a negative outer power

Differentiate y=(x2+1)−3y=(x^2+1)^{-3}.

Plan. Apply the power rule to the whole bracket, then multiply by 2x2x.

dydx=−3(x2+1)−4×2x=−6x(x2+1)−4.\frac{dy}{dx}=-3(x^2+1)^{-4}\times2x=-6x(x^2+1)^{-4}.

Answer. −6x/(x2+1)4-6x/(x^2+1)^4.

Check. At x=1x=1, the result is −3/8-3/8. A central secant with step 0.00010.0001 gives approximately −0.375000-0.375000; the function is decreasing for positive xx, agreeing with the sign.

The inner factor is essential. 3(x2+1)23(x^2+1)^2 alone is a derivative with respect to uu, not with respect to xx.

The product rule

When two factors both change, each contributes to the change in their product.

Imagine a rectangle with side lengths uu and vv. Changing them by small amounts Δu\Delta u and Δv\Delta v changes its area by

(u+Δu)(v+Δv)−uv=vΔu+uΔv+ΔuΔv.(u+\Delta u)(v+\Delta v)-uv =v\Delta u+u\Delta v+\Delta u\Delta v.

After dividing by the input change and taking the derivative limit, the final tiny product term disappears. The two main contributions remain.

[DIAGRAM: asset_name: Two contributions to a changing product; asset_slug: product-area; caption: Each changing side adds a strip; both strips contribute to the product rule.; description: A rectangle of width u and height v is extended by positive increments delta u and delta v. The right strip is labelled v delta u, the top strip u delta v and the small corner delta u delta v. Brackets label the original and added side lengths.]
Each changing side adds a strip

Product rule

For differentiable functions u(x)u(x) and v(x)v(x),

ddx(uv)=u′v+uv′.\frac{d}{dx}(uv)=u'v+uv'.

Differentiate one factor at a time and leave the other unchanged. Not in the formula booklet: learn it.

A numerical coefficient is constant, so it can stay outside the calculation. A sum of functions uses separate derivatives; a product requires this two-term rule.

Worked example 3: a polynomial times a sine

Differentiate y=x2sin⁡xy=x^2\sin x.

Plan. Take u=x2u=x^2, v=sin⁡xv=\sin x, then add their two contributions.

u′=2x,v′=cos⁡x,dydx=(2x)sin⁡x+x2cos⁡xapply u′v+uv′.\begin{aligned} u'&=2x,&v'&=\cos x,\\ \frac{dy}{dx}&=(2x)\sin x+x^2\cos x &&\text{apply }u'v+uv'. \end{aligned}

Answer. 2xsin⁡x+x2cos⁡x2x\sin x+x^2\cos x.

Check. At x=πx=\pi, the first contribution is zero and the second is −π2-\pi^2. Near π\pi, the positive factor x2x^2 multiplies a sine changing from positive to negative, so a negative gradient makes sense.

Change only the polynomial exponent from 22 to 33. Both contributions must be updated.

Worked example 4: a different power factor

Differentiate y=x3sin⁡xy=x^3\sin x.

Plan. Keep the sine unchanged in the first term and the whole cubic unchanged in the second.

dydx=3x2sin⁡x+x3cos⁡x.\frac{dy}{dx}=3x^2\sin x+x^3\cos x.

Answer. 3x2sin⁡x+x3cos⁡x3x^2\sin x+x^3\cos x.

Check. At x=1x=1, the derivative is 3sin⁡1+cos⁡1≈3.0647153\sin1+\cos1\approx3.064715. A central secant with step 0.00010.0001 gives approximately 3.0647153.064715.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account