There are certain ‘rules’ we can use when trying to find the Derivative of a Function, especially when the function is some composite. Depending on the form of the function to be differentiated, we can use these rules to break down the differentiation process to make it easier to solve analytically.
Power rule
The power rule states that if we have a function that is a variable raised to some power, the derivative is equal to the power multiplied by the variable raised to that power minus one.
Remark
Lagrange’s notation:
\begin{align} h(x) &= x^n \ h’(x) &= n \cdot x^{n-1} \end{align}
\begin{align} y &= x^n \ \frac{dy}{dx} &= n \cdot x^{n-1} \end{align}
Power rule
Applying the power rule:
\begin{align} h’(x) &= 5 \cdot x^{5-1} \ &= 5x^4 \end{align}
> [!example]- Power rule with coefficient > > $h(x) = 3x^4$ > > The coefficient can be brought along when differentiating: > $$ \begin{align} h'(x) &= 3 \cdot 4 \cdot x^{4-1} \\ &= 12x^3 \end{align}This rule forms the basis for many other differentiation techniques and is one of the most fundamental rules in calculus. It works for any real number power, including negative and fractional powers, though the derivation becomes more complex in those cases.
Sum Rule
The sum rule states that if we have a function that is the sum (linear combination) of some other functions, the total derivative of the function is equal to the sum of the derivatives.
Remark
Lagrange’s notation:
\begin{align} h(x) &= A,f(x) + B, g(x) \ h’(x) &= A,f’(x) + B,g’(x) \end{align}
\begin{align} y &= A,f(x) + B,g(x) \ \frac{dy}{dx} &= A \frac{df}{dx} + B \frac{dg}{dx} \end{align}
## Product Rule The product rule states that if we have a function that is the product of some other functions, the derivative of the function is equal to the sum of the product of the derivative of each RHS function with the other RHS functions > [!remark|*] > > Lagrange's notation: > $$ \begin{align} h(x) &= f(x) \cdot g(x) \\ h'(x) &= f'(x) \cdot g(x) + f(x) \cdot g'(x) \end{align} > $$ > Leibniz's notation: > $$ \begin{align} y &= f(x) \cdot g(x) \\ \frac{dy}{dx} &= \frac{df}{dx} \cdot g + \frac{dg}{dx} \cdot f \end{align}Chain Rule
The chain rule states that if we have a function that is a composite function of two functions, it’s derivative is the product of the derivative of the outer function and derivative of the inner function.
Remark
Lagrange’s notation:
\begin{align} h(x) &= f(g(x)) \ h’(x) &= f’(g(x)) \cdot g’(x) \end{align}
\begin{align} y &= f(g(x)) \ y &= f(u), \ u = g(x) \ \frac{dy}{dx} &= \frac{dy}{du} \cdot \frac{du}{dx} \end{align}
> [!example]- Chain rule > > $h(x) = (x^2 + 1)^3 = f(g(x)^3)$ > where $g(x) = x^2 + 1$. > > First let's find the derivative of $f$ wr.t. $g(x)$ i.e. $f^\prime(g(x))$. > $$\begin{align} f(g(x)) = g(x)^3 \\ f'(g(x)) = 3g(x)^2 \end{align} > $$ > Then let's find the derivative of $g$ w.r.t $x$ i.e. $g'(x)$. > $$ \begin{align} g(x) = x^2 + 1 \\ g'(x) = 2x \end{align} > $$ > Applying the chain rule we get: $$ \begin{align} h'(x) = 3(x^2 + 1)^2 \cdot 2x \\ h'(x) = 6x \cdot (x^2 + 1)^2 \end{align}Chain rule
\begin{align} \frac{dy}{dx} &= \frac{dy}{du} \frac{du}{dx} \ \frac{dy}{dx} &= \frac{d}{du}\sin(u) \cdot \frac{d}{dx} x^2 \ &= \cos(x^2) \cdot 2x \end{align}
The chain rule can be understood through rates of change: if you know how $z$ changes with respect to $y$, and how $y$ changes with respect to $x$, you can find the rate $z$ changes with respect to $x$ by taking the product these two rates. Physical intuition: if a train moves $3$ times [[Velocity|faster]] than a car, and a car moves $5$ times faster than a bike, then the train must move $3 × 5 = 15$ times faster than the bike. The rates of change compound through multiplication, just like in this simple speed comparison. ## Quotient rule The quotient rule states that if we have a function that is the quotient of two other functions, the derivative is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. > [!remark|*] > > Lagrange's notation: > $$ \begin{align} h(x) &= \frac{f(x)}{g(x)} \\ h'(x) &= \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \end{align} > $$ > Leibniz's notation: > $$ \begin{align} y &= \frac{f(x)}{g(x)} \\ \frac{dy}{dx} &= \frac{g\frac{df}{dx} - f\frac{dg}{dx}}{g^2} \end{align}Quotient rule
where and
First find the derivatives of and :
\begin{align} f’(x) &= 2x \ g’(x) &= 1 \end{align}
\begin{align} h’(x) &= \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2} \ &= \frac{2x^2 + 2x - x^2}{(x+1)^2} \ &= \frac{x^2 + 2x}{(x+1)^2} \end{align}
## Reciprocal rule The reciprocal rule states that the derivative of the reciprocal of a function is equal to the negative of the derivative of the function divided by the square of the function. > [!remark|*] > > Lagrange's notation: > $$ \begin{align} h(x) &= \frac{1}{f(x)} = f(x)^{-1} \\ h'(x) &= -\frac{f'(x)}{f(x)^2} \end{align} > $$ > Leibniz's notation: > $$ \begin{align} y &= \frac{1}{f(x)} = f(x)^{-1} \\ \frac{dy}{dx} &= -\frac{1}{f(x)^2} \cdot \frac{df}{dx} \end{align}Reciprocal rule
Applying the reciprocal rule:
\begin{align} h’(x) &= -\frac{2x}{(x^2)^2} \ &= -\frac{2x}{x^4} \ &= -\frac{2}{x^3} \end{align}
> [!note]- Derivation from Power and Chain Rules > > We can derive the reciprocal rule using the power rule and chain rule: > > Let $h(x) = f(x)^{-1}$ > > Using the chain rule: > $$ \begin{align} h'(x) &= -1 \cdot f(x)^{-2} \cdot f'(x) \\ &= -\frac{f'(x)}{f(x)^2} \end{align} > $$ > > This works because: > 1. The power rule tells us that $\frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2}$ > 2. The chain rule tells us to multiply by the derivative of the inner function $f'(x)$ > 3. Combining these gives us the reciprocal rule formula This rule is particularly useful when differentiating expressions involving fractions where the numerator is 1, and can be seen as a special case of both the power rule and the quotient rule. ## Exponential and logarithmic rules For any [[Exponentiation#^5f7806|exponential]] function the derivative can be defined as: > [!remark|*] > > Lagrange's notation: > $$ \begin{align} h(x) &= b^{ax} \\ h'(x) &= a \cdot b^{ax} \cdot \ln(b) \end{align} > $$ > Leibniz's notation: > $$ \begin{align} \\ y &= b^{ax} \\ \frac{dy}{dx} &= a \cdot b^{ax} \cdot \ln (b) \end{align} > $$ ^fb7c76 From this we can see that the derivative of the natural exponential function $e^x$ is itself.\begin{align} f(x) &= e^{a x} \ \
f’(x) &= a \cdot e^{a x} \cdot \ln (e) \ &= a \cdot e^{a x} \cdot 1 \ &= a,e^{a x} \end{align}
\frac{d}{dx} \log_{c}(x) = \frac{1}{x \ln(c)}.
## Inverse-function viewpoint Many rules become easier to remember by pairing them with inverse functions. [[Root]] functions use fractional powers, so the power rule applies. [[Logarithm|Logarithms]] invert exponentials, so their derivatives inherit the reciprocal shape $1/x$. If $y=f^{-1}(x)$, then\frac{dy}{dx}=\frac{1}{f’(y)},
where the derivative exists and $f'(y)\ne0$. ## Common pitfalls - Forgetting the inner derivative in the [[#chain-rule|chain rule]]. - Applying the power rule to $a^x$; exponentials need the factor $\ln a$. - Treating $\frac{d}{dx}\ln(g(x))$ as $1/g(x)$ instead of $g'(x)/g(x)$. - Using the quotient rule when a simpler rewrite with negative powers would be cleaner.