The product rule is a picture
Tonight's idea
You know the derivative of \(x^2\) is \(2x\) and of \(x^3\) is \(3x^2\) (grade 10). What about a product of two functions, \(u(x)\,v(x)\)? The tempting guess \(u'v'\) is wrong: \(x \cdot x = x^2\) would give \(1 \cdot 1 = 1\), not \(2x\). The right rule comes from a rectangle.
Draw a rectangle with sides \(u\) and \(v\), so its area is \(uv\). Now nudge \(x\) by a small amount \(\Delta x\) (read: a change in \(x\)). The sides grow by \(\Delta u\) and \(\Delta v\), and the area gains three pieces:
So the change in the product is \(\Delta(uv) = v\,\Delta u + u\,\Delta v + \Delta u\,\Delta v\). Divide by \(\Delta x\) to get a rate:
\[\frac{\Delta(uv)}{\Delta x} = v\,\frac{\Delta u}{\Delta x} + u\,\frac{\Delta v}{\Delta x} + \Delta u \cdot \frac{\Delta v}{\Delta x}.\]
Now let \(\Delta x \to 0\). The two ratios become the derivatives \(u'\) and \(v'\). The last term is \(\Delta u\) times something tending to \(v'\), and \(\Delta u\) itself tends to 0. The corner contributes nothing.
\((uv)' = u'v + uv'\). Differentiate each factor in turn, leave the other alone, and add.
The corner is the lesson: a product of two small changes is too small for a derivative to see. Check on \(x \cdot x\): \(1 \cdot x + x \cdot 1 = 2x\). Correct this time.
Watch one: \(y = (x^2 + 1)(x^3 - x)\), two ways
Product rule, with \(u = x^2 + 1\) and \(v = x^3 - x\):
\[y' = 2x\,(x^3 - x) + (x^2 + 1)(3x^2 - 1) = 2x^4 - 2x^2 + 3x^4 + 2x^2 - 1 = 5x^4 - 1.\]
Expanding first: \(y = x^5 - x\), so \(y' = 5x^4 - 1\). The two routes agree, but the point is that the rule works when you cannot expand, as with \(x^2 \sin x\) or \(x e^x\) once you have the derivatives of \(\sin x\) and \(e^x\) (sheet 4). The picture also fixes the sign: both strips are added, never subtracted.
See a derivative as the limit of secant slopes, which is what \(\Delta u / \Delta x \to u'\) means: mathdetours.com/demos/tangent.html.
Your turn
- Using only \((x^2)' = 2x\) and the product rule, differentiate \(x^3 = x \cdot x^2\). Then differentiate \(x^4 = x \cdot x^3\).
- Differentiate \(y = (2x + 1)(x^2 - 3)\) by the product rule, then by expanding first. Then differentiate \(y = x^2 \cdot \dfrac1x\) by the product rule, using \(\left(\dfrac1x\right)' = -\dfrac1{x^2}\), and explain why the answer had to be 1.
- Two functions have \(f(2) = 3\), \(f'(2) = -1\), \(g(2) = 5\), \(g'(2) = 4\). Find (a) the derivative of \(f(x)\,g(x)\) at \(x = 2\), and (b) the derivative of \(f(x)^2\) at \(x = 2\), by writing \(f^2 = f \cdot f\). What is the general formula for \((f^2)'\)?
- Prove the power rule \((x^n)' = n x^{n-1}\) for every positive integer \(n\) by induction: it holds for \(n = 1\), and if it holds for \(n\), apply the product rule to \(x^{n+1} = x \cdot x^n\). Then describe the picture for three factors and write down \((uvw)'\).
Stretch: the quotient rule for free
Let \(q = u/v\), so \(u = q\,v\). Differentiate both sides with the product rule, solve for \(q'\), and tidy it into a single fraction over \(v^2\). Use it to differentiate \(\dfrac{x}{x^2 + 1}\).
Answers
- \((x \cdot x^2)' = 1 \cdot x^2 + x \cdot 2x = 3x^2\). \((x \cdot x^3)' = 1 \cdot x^3 + x \cdot 3x^2 = 4x^3\).
- \(2(x^2 - 3) + (2x + 1)(2x) = 6x^2 + 2x - 6\); expanding, \(y = 2x^3 + x^2 - 6x - 3\), with the same derivative. \(\left(x^2 \cdot \tfrac1x\right)' = 2x \cdot \tfrac1x + x^2 \cdot \left(-\tfrac1{x^2}\right) = 2 - 1 = 1\), as it must, since \(x^2 \cdot \tfrac1x = x\).
- (a) \(f'g + fg' = (-1)(5) + (3)(4) = 7\). (b) \(f'f + ff' = 2ff' = 2(3)(-1) = -6\). In general \((f^2)' = 2ff'\), a first glimpse of the chain rule (next sheet).
- Base case: \((x^1)' = 1 = 1 \cdot x^0\). Step: \((x \cdot x^n)' = 1 \cdot x^n + x \cdot n x^{n-1} = (n + 1)x^n\), which is the rule for \(n + 1\). For three factors the picture is a box \(u \times v \times w\) gaining three slabs, three thin rods and one tiny corner cube; only the slabs survive: \((uvw)' = u'vw + uv'w + uvw'\). Equivalently, apply the two-factor rule to \((uv)\cdot w\).
Stretch. \(u' = q'v + qv'\), so \(q' = \dfrac{u' - qv'}{v} = \dfrac{u' - \frac{u}{v}v'}{v} = \dfrac{u'v - uv'}{v^2}\). For \(\dfrac{x}{x^2 + 1}\): \(u' = 1\), \(v' = 2x\), so \(q' = \dfrac{(x^2 + 1) - x \cdot 2x}{(x^2 + 1)^2} = \dfrac{1 - x^2}{(x^2 + 1)^2}\). It is zero at \(x = \pm 1\), where the curve has its highest and lowest points.