The same, in symbols

\delta(uv) = u\,\delta v + v\,\delta u + \delta u\,\delta v

\frac{\delta(uv)}{\delta x} = u\frac{\delta v}{\delta x} + v\frac{\delta u}{\delta x} + \delta u\,\frac{\delta v}{\delta x}

\delta x \to 0\!: \quad \delta u \to 0

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

the corner term is δu times something finite, and δu goes to zero

Differentiation rules

To differentiate a product, grow a rectangle.

The derivative of a product is not the product of the derivatives. Picture the product as an area and the right rule draws itself.

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A product is an area

Think of $y = uv$ as the area of a rectangle with sides $u$ and $v$ . Here $u = 1 + x$ and $v = 1 + \tfrac{1}{2}x^2$ , and at $x = 1$ the sides are $2$ and $1.5$ .

Three new pieces

When $x$ grows by $\delta x$ , $u$ grows by $\delta u$ and $v$ by $\delta v$ . The area gains a strip $v\,\delta u$ down the side, a strip $u\,\delta v$ along the top, and a corner $\delta u\,\delta v$ .

The corner vanishes

Divide each piece by $\delta x$ and shrink $\delta x$ . The strips settle on $1.5$ and $2$ . The corner is one small number times another, and its share goes to $0$ . The rate settles on $3.5$ , and in general:

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

Using it

Example 12.5: for $y = x^2e^{3x}$ , take $u = x^2$ and $v = e^{3x}$ . Then

\begin{aligned} \frac{dy}{dx} &= 3x^2e^{3x} + 2xe^{3x} \\ &= xe^{3x}(3x + 2) \end{aligned}

so the stationary points are at $x = 0$ and $x = -\tfrac{2}{3}$ .

Chapter 12

Two strips count; the corner never does.

sin and cos from first principles, eˣ and ln x, the chain rule, the quotient rule, and twelve graded exam questions with full worked solutions are waiting.

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