Coaching Centre
The product rule
Differentiating a product of two functions with (uv)′ = u′v + uv′, combining it with the chain rule, and factorising the result.
The product rule
Differentiate $y = x^3(2x - 1)^4$ and factorise.
For $y = x(x - 4)^3$, find where $\dfrac{dy}{dx} = 0$.
If $y = uv$, then $\dfrac{dy}{dx} = $
That is the product rule.
For $y = x^2(x + 1)^3$, find $\dfrac{dy}{dx}$ at $x = 1$.
$2x(x + 1)^3 + 3x^2(x + 1)^2 = 16 + 12 = 28$.
$\dfrac{d}{dx}\,x^3(2x - 1)^4$ factorises to:
See Example 1.
For $y = x\sqrt{x}$, find $\dfrac{dy}{dx}$ at $x = 4$.
$y = x^{\frac32}$, so $\dfrac{dy}{dx} = \tfrac32\sqrt{x} = 3$. The product rule gives the same answer.
For $y = (x + 1)(x^2 - 2)$, find $\dfrac{dy}{dx}$ at $x = 2$.
$(x^2 - 2) + (x + 1)(2x) = 2 + 12 = 14$.
Which check shows that $\dfrac{d}{dx}(uv) \ne u'v'$?
A single counterexample is enough.
For $y = x(3x + 2)^2$, find $\dfrac{dy}{dx}$ at $x = 0$.
$(3x + 2)^2 + 6x(3x + 2) = 4$ at $x = 0$.
For $y = x(x - 4)^3$, find both $x$-values where $\dfrac{dy}{dx} = 0$.
$(x - 4)^2(4x - 4) = 0$.
For $f(x) = (2x + 1)(x - 3)^2$, find $f'(0)$.
$2(x - 3)^2 + 2(2x + 1)(x - 3) = 18 - 6 = 12$.
New numbers every time. Keep going until you get five in a row.