Product, Quotient and Chain Rules

The product of two functions does not differentiate to the product of their derivatives. Each function takes a turn being differentiated while the other holds still.

Always start by identifying $u = f(x)$ and $v = g(x)$, writing their derivatives, then substituting into $u'v + uv'$.

Product rule: $(fg)' = f'g + fg'$, "derivative of first times second, plus first times derivative of second."; Step: label $u$ and $v$, find $u'$ and $v'$, substitute. Don't skip the labelling step.

Pause, copy the product rule $(fg)' = f'g + fg'$, "derivative of first times second, plus first times derivative of second", and the labelling method (always name $u$ and $v$ first) into your book.

We just saw the product rule $(fg)' = f'g + fg'$ for multiplied functions. That raises a question: when one function divides another, is there an analogous rule, and does the order of $f$ and $g$ matter? This card answers it → the quotient rule $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$, where order matters because of the subtraction in the numerator.

When one function divides another, the quotient rule handles the differentiation, with a crucial subtraction in the numerator and the denominator squared on the bottom.

Quotient rule: $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$, the order matters: it's subtraction, not addition.; Mnemonic: "lo d(hi) minus hi d(lo), over lo squared."

Pause, copy the quotient rule $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$ and the mnemonic "lo d(hi) minus hi d(lo), over lo squared" into your book.

We just saw the product and quotient rules, both handle two functions side-by-side. That raises a question: what rule applies when one function is plugged inside another, such as $\ln(\cos x)$ or $e^{\sin x}$? This card answers it → the chain rule $[f(g(x))]' = f'(g(x)) \cdot g'(x)$: differentiate outside (leaving inside untouched), then multiply by the inside's derivative.

The chain rule applies whenever one function is plugged inside another, a composite. The key step is identifying what is "outside" and what is "inside," then multiplying the two derivatives.

Chain rule: $[f(g(x))]' = f'(g(x)) \cdot g'(x)$, differentiate outside (leave inside alone), multiply by inside derivative.; $\ln(\cos x) \Rightarrow y' = -\tan x$, worth memorising as a result.

Pause, copy the chain rule $[f(g(x))]' = f'(g(x)) \cdot g'(x)$ and the key result $\dfrac{d}{dx}\ln(\cos x) = -\tan x$ into your book.

We just saw the three differentiation rules, product, quotient, and chain, for computing $f'(x)$. That raises a question: once you have the derivative at a specific point, how do you write the equation of the tangent or normal line to the curve? This card answers it → evaluate $m_T = f'(x_0)$, then apply point-slope form $y - y_0 = m_T(x - x_0)$; for the normal use $m_N = -1/m_T$, which needs $m_T \ne 0$.

The derivative gives the exact slope of the curve at any point. Once you have the slope, writing the equation of the tangent or normal is just point-slope form.

Tangent at $(x_0, y_0)$: find $m_T = f'(x_0)$, then use $y - y_0 = m_T(x-x_0)$.; Normal at $(x_0, y_0)$: $m_N = -1/m_T$, same point-slope form.

Pause, copy the tangent formula $y - y_0 = m_T(x-x_0)$ where $m_T = f'(x_0)$ and the normal slope $m_N = -1/m_T$, valid when $m_T \ne 0$ (if $m_T = 0$ the normal is $x = x_0$), into your book.

We just saw the tangent and normal method: differentiate to find $m_T$, then apply point-slope form. That raises a question: in HSC questions like $y = x^2e^{-x}$ or $y = \dfrac{\sin x}{1+\cos x}$, which rule do you reach for first, and how do you combine product, quotient, and chain in the right order? This card answers it → always identify the outermost structure first, apply that rule, then handle each component with its own rule.

The hardest HSC problems combine rules in a single expression. The strategy is always the same: identify the outermost structure first (product, quotient, or composition), apply the corresponding rule, then handle each component using its own rule.

Strategy: identify outermost structure (product/quotient/composite), apply that rule first.; $y = x^2 e^{-x}$: product rule gives $y' = xe^{-x}(2-x)$. Stationary at $x=0$ and $x=2$.

Pause, copy the strategy (identify outermost structure first, then apply product/quotient/chain) and the worked result $\dfrac{d}{dx}[x^2e^{-x}] = xe^{-x}(2-x)$ with stationary points at $x = 0$ and $x = 2$ into your book.