For any function $f(x) = x^n$ where $n$ is any real number:

$\frac{d}{dx}(x^n) = nx^{n-1}$

Example 2: Differentiate $f(x) = x^5$

Solution:

$f'(x) = 5x^{5-1} = 5x^4$

For sums and differences:

$\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$

$\frac{d}{dx}[kf(x)] = kf'(x) \text{ (where k is constant)}$

When $y$ is a function of $u$, and $u$ is a function of $x$:

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

Example 3: Differentiate $y = (2x^2 + 4)^6$

Solution:

Let $u = 2x^2 + 4$, then $y = u^6$

$\frac{du}{dx} = 4x, \quad \frac{dy}{du} = 6u^5$

$\frac{dy}{dx} = 6u^5 \times 4x = 24x(2x^2 + 4)^5$

If $y = u(x) \cdot v(x)$, then:

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

Example 4: Differentiate $f(x) = (x^3 - 2)(x+4)^2$

Solution:

Let $u = x^3 - 2$ and $v = (x+4)^2$

$\frac{du}{dx} = 3x^2, \quad \frac{dv}{dx} = 2(x+4)$

$f'(x) = (x+4)^2(3x^2) + (x^3 - 2)(2x+8)$

If $y = \frac{u}{v}$, then:

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

Example 5: Differentiate $y = \frac{x^2 + 4}{x^2 - 4}$

Solution:

Let $u = x^2 + 4$ and $v = x^2 - 4$

$\frac{du}{dx} = 2x, \quad \frac{dv}{dx} = 2x$

$\frac{dy}{dx} = \frac{(x^2 - 4)(2x) - (x^2 + 4)(2x)}{(x^2 - 4)^2} = \frac{-16x}{(x^2 - 4)^2}$

For functions of two variables $z = f(x, y)$, partial derivatives measure the rate of change with respect to one variable while holding the other constant.

Partial derivative with respect to x:

$\frac{\partial z}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h}$

Partial derivative with respect to y:

$\frac{\partial z}{\partial y} = \lim_{h \to 0} \frac{f(x, y+h) - f(x, y)}{h}$

Example 6: Find the first partial derivatives of $z = x^2 + 3xy + y^2$

Solution:

$\frac{\partial z}{\partial x} = 2x + 3y$

$\frac{\partial z}{\partial y} = 3x + 2y$

Second partial derivatives are obtained by differentiating partial derivatives again:

$\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)$

$\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)$

In Tanzania, differentiation is used in many practical situations. For example, a small business owner in Dar es Salaam calculating profit margins can use derivatives to determine the rate of change of profit with respect to the number of items sold, helping them find the optimal pricing. Engineers at construction sites use differentiation to calculate rates of change, such as determining how fast water levels change in storage tanks, which is essential for designing water supply systems in communities.