Differential Calculus: Simple and Partial Derivatives

Techniques of Differentiation

1. Constant rule

The derivative of a constant term is zero. If $y = c$ (a constant), then

$$\frac{dy}{dx} = 0$$

2. Power rule

Differentiating $y = x^n$ with respect to $x$ gives

$$\frac{dy}{dx} = n x^{n-1}$$

For example, for $y = x$,

$$ \begin{aligned} \frac{dy}{dx} &= 1 \cdot x^{1-1} \ &= 1 \cdot x^0 \ &= 1 \end{aligned} $$

For $y = 6x^2$,

$$ \begin{aligned} \frac{dy}{dx} &= 2 \cdot 6 x^{2-1} \ &= 12x \end{aligned} $$

3. Sum rule

If $y = u + v$, then

$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$

For example, let $y = x^5 + x^9$. Differentiating term by term,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(x^5) + \frac{d}{dx}(x^9) \ = 5x^{5-1} + 9x^{9-1} \ = 5x^4 + 9x^8 \end{aligned} $$

4. Product rule

If $y = uv$, then

$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$

For example, let $y = x^4(x^3 + 2)$, so $u = x^4$ and $v = x^3 + 2$. Applying the product rule,

$$ \begin{aligned} \frac{dy}{dx} &= x^4 \frac{d}{dx}(x^3 + 2) + (x^3 + 2)\frac{d}{dx}(x^4) \ = x^4 \cdot 3x^2 + (x^3 + 2)\cdot 4x^3 \ = 3x^6 + 4x^6 + 8x^3 \ = 7x^6 + 8x^3 \end{aligned} $$

5. Quotient rule

If $y = \dfrac{u}{v}$, then

$$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$

For example, let $y = \dfrac{x^3 + 2}{x}$, so $u = x^3 + 2$ and $v = x$. Applying the quotient rule,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{x \cdot \frac{d}{dx}(x^3 + 2) - (x^3 + 2)\cdot \frac{d}{dx}(x)}{x^2} \ = \frac{x \cdot 3x^2 - (x^3 + 2)\cdot 1}{x^2} \ = \frac{3x^3 - x^3 - 2}{x^2} \ = \frac{2x^3 - 2}{x^2} \end{aligned} $$

6. Chain rule

If $y = f(u)$ and $u = g(x)$, then

$$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$$

For example, let $y = u^5$ and $u = x^4 + x + 2$. Differentiating each part,

$$ \begin{aligned} \frac{dy}{du} &= 5u^4 \ \frac{du}{dx} &= 4x^3 + 1 \end{aligned} $$

Combining them,

$$ \begin{aligned} \frac{dy}{dx} &= 5u^4 (4x^3 + 1) \ &= 5(x^4 + x + 2)^4 (4x^3 + 1) \end{aligned} $$

Partial Derivatives (First Order)

When a function depends on more than one variable, a partial derivative differentiates with respect to one variable while treating the others as constants. For $u = f(x, y)$, the partial derivative $u_x$ treats $y$ as constant, and $u_y$ treats $x$ as constant.

For example, find $u_x$ and $u_y$ if $u = 3x^3 + 4xy + y^2$.

Differentiating with respect to $x$ (treating $y$ as constant),

$$ \begin{aligned} u_x &= \frac{\partial u}{\partial x} \ &= 9x^2 + 4y \end{aligned} $$

Differentiating with respect to $y$ (treating $x$ as constant),

$$ \begin{aligned} u_y &= \frac{\partial u}{\partial y} \ &= 4x + 2y \end{aligned} $$