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Differentiation Rules

Differentiation Rules

Here’s a handy summary of the differentiation rules you’ll frequently use.


Product Rule

The differentiation rule for the product of two functions:
\begin{align*}
(fg)’&= f’g + fg’\\[8px]
&= [{\small \text{ (deriv of the 1st) } \times \text{ (the 2nd) }}]\, + \,[{\small \text{ (the 1st) } \times \text{ (deriv of the 2nd)}}]
\end{align*}

• For examples of the Product Rule, visit our Calculating Derivatives: Problems & Solutions page!

Quotient Rule

The differentiation rule for the quotient of two functions:
\begin{align*}
\dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px]
&=\dfrac{{[{\small \text{(deriv of numerator) } \times \text{ (denominator)}]}\\ \quad – \, [{\small \text{ (numerator) } \times \text{ (deriv of denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}}
\end{align*}
Many students remember the quotient rule by thinking of the numerator as “hi,” the demoninator as “lo,” the derivative as “d,” and then singing

“lo d-hi minus hi d-lo over lo-lo”

• For examples of the Quotient Rule: Calculating Derivatives: Problems & Solutions.

Chain Rule

The differentiation rule for the composition of two functions:
\begin{align*}
\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[8px]
&= \text{[derivative of the outer function, evaluated at the inner function]}\\[8px]
& \qquad \times \text{[derivative of the inner function]}
\end{align*}


Alternatively, if we write $y = f(u)$ and $u = g(x),$ then
$$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} $$

Informally:
$$\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}$$

• For many examples of the Chain Rule: Chain Rule: Problems & Solutions.



One quick example: Consider $f(x) = (x^2 + 1)^7.$
To find the derivative, think something like: “The function is a bunch of stuff to the 7th power. So the derivative is 7 times that same stuff to the 6th power, times the derivative of that stuff.”

\begin{align*}
f(x) &= (\text{stuff})^7; \quad \text{stuff} = x^2 + 1 \\[12px]
\text{Then}\phantom{f(x)= }\\
\frac{df}{dx} &= 7(\text{stuff})^6 \cdot \left(\frac{d}{dx}(x^2 + 1)\right) \\[8px]
&= 7(x^2 + 1)^6 \cdot (2x) \quad \cmark
\end{align*}
Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives.


Tip: You can differentiate any function, for free,
using Wolfram WolframAlpha’s Online Derivative Calculator.