Practice Questions: Differentiation from First Principles

$$\begin{align*} g'(x) &= \lim_{h \to 0}\frac{g(x+h)-g(x)}{h} \\ &= \lim_{h \to 0}\frac{2(x+h)-3-(2x-3)}{h} \\ &= \lim_{h \to 0}\frac{2x+2h-3-2x+3}{h} \\ &= \lim_{h \to 0}\frac{2h}{h} \\ &= \lim_{h \to 0} 2 \\ &= 2 \end{align*}$$

$$\begin{align*} f'(x) &= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0}\frac{4(x+h)^3-4x^3}{h} \\ &= \lim_{h \to 0}\frac{4(x^3+3x^2h+3xh^2+h^3)-4x^3}{h} \\ &= \lim_{h \to 0}\frac{4x^3+12x^2h+12xh^2+4h^3-4x^3}{h} \\ &= \lim_{h \to 0}\frac{12x^2h+12xh^2+4h^3}{h} \\ &= \lim_{h \to 0}\frac{h(12x^2+12xh+4h^2)}{h} \\ &= \lim_{h \to 0}(12x^2+12xh+4h^2) \\ &= 12x^2 \end{align*}$$

$$\begin{align*} \frac{dp}{dx} &= \lim_{h \to 0}\frac{p(x+h)-p(x)}{h} \\ &= \lim_{h \to 0}\frac{-\frac{2}{x+h}-(-\frac{2}{x})}{h} \\ &= \lim_{h \to 0}\frac{1}{h}\left(\frac{-2}{x+h}+\frac{2}{x}\right) \\ &= \lim_{h \to 0}\frac{1}{h}\left(\frac{-2x+2(x+h)}{x(x+h)}\right) \\ &= \lim_{h \to 0}\frac{1}{h}\left(\frac{-2x+2x+2h}{x(x+h)}\right) \\ &= \lim_{h \to 0}\frac{1}{h}\left(\frac{2h}{x^2+xh}\right) \\ &= \lim_{h \to 0}\frac{2}{x^2+xh} \\ &= \frac{2}{x^2} \end{align*}$$

$$\begin{align*} g'(x) &= \lim_{h \to 0}\frac{g(x+h)-g(x)}{h} \\ &= \lim_{h \to 0}\frac{\frac{1}{4}-\frac{1}{4}}{h} \\ &= \lim_{h \to 0}\frac{0}{h} \\ &= \lim_{h \to 0} 0 \\ &= 0 \end{align*}$$

Interpretation: The derivative of a constant function is always zero. This means the gradient of the function $g(x) = \frac{1}{4}$ is zero at any point on the graph, indicating a horizontal line.

$$\begin{align*} g(x) &= -x^2 \\ g(x+h) &= -(x+h)^2 \\ \therefore \frac{g(x+h)-g(x)}{h} &= \frac{-(x+h)^2-(-x^2)}{h} \\ &= \frac{-(x^2+2xh+h^2)+x^2}{h} \\ &= \frac{-x^2-2xh-h^2+x^2}{h} \\ &= \frac{-2xh-h^2}{h} \\ &= \frac{h(-2x-h)}{h} \\ &= -2x-h \end{align*}$$

$$\begin{align*} \lim_{h \to 0}\frac{g(x+h)-g(x)}{h} &= \lim_{h \to 0}(-2x-h) \\ &= -2x-0 \\ &= -2x \end{align*}$$

Explanation: The derivative of $g(x)$ is $g'(x) = -2x$. This means the gradient of the function $g$ is given by the expression $-2x$. The gradient of the graph depends on the value of $x$ - it is negative when $x > 0$ and positive when $x < 0$.

$$\begin{align*} f(x) &= -2x^2+3x+1 \\ f(x+h) &= -2(x+h)^2+3(x+h)+1 \\ f'(x) &= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0}\frac{-2(x+h)^2+3(x+h)+1-(-2x^2+3x+1)}{h} \\ &= \lim_{h \to 0}\frac{-2(x^2+2xh+h^2)+3x+3h+1+2x^2-3x-1}{h} \\ &= \lim_{h \to 0}\frac{-2x^2-4xh-2h^2+3h+2x^2}{h} \\ &= \lim_{h \to 0}\frac{-4xh-2h^2+3h}{h} \\ &= \lim_{h \to 0}\frac{h(-4x-2h+3)}{h} \\ &= \lim_{h \to 0}(-4x-2h+3) \\ f'(x) &= -4x+3 \end{align*}$$

$$\begin{align*} f(x) &= \frac{1}{x-2} \\ f(x+h) &= \frac{1}{x+h-2} \\ f'(x) &= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0}\frac{\frac{1}{x+h-2}-\frac{1}{x-2}}{h} \\ &= \lim_{h \to 0}\frac{\frac{(x-2)-(x+h-2)}{(x+h-2)(x-2)}}{h} \\ &= \lim_{h \to 0}\frac{\frac{x-2-x-h+2}{(x+h-2)(x-2)}}{h} \\ &= \lim_{h \to 0}\left(\frac{-h}{(x+h-2)(x-2)}\right) \times \frac{1}{h} \\ &= \lim_{h \to 0}\frac{-1}{(x+h-2)(x-2)} \\ f'(x) &= \frac{-1}{(x-2)^2} \end{align*}$$

$$\begin{align*} g(x) &= -5x^2 \\ g(x+h) &= -5(x+h)^2 \\ g'(x) &= \lim_{h \to 0}\frac{g(x+h)-g(x)}{h} \\ &= \lim_{h \to 0}\frac{-5(x^2+2xh+h^2)-(-5x^2)}{h} \\ &= \lim_{h \to 0}\frac{-5x^2-10xh-5h^2+5x^2}{h} \\ &= \lim_{h \to 0}\frac{-10xh-5h^2}{h} \\ &= \lim_{h \to 0}\frac{h(-10x-5h)}{h} \\ &= \lim_{h \to 0}(-10x-5h) \\ &= -10x \end{align*}$$

Therefore:

$$\begin{align*} g'(3) &= -10(3) \\ &= -30 \end{align*}$$

First, expand $p(x)$:

$$\begin{align*} p(x) &= 4x(x-1) \\ &= 4x^2-4x \end{align*}$$

Now find the derivative:

$$\begin{align*} p(x+h) &= 4(x+h)^2-4(x+h) \\ p'(x) &= \lim_{h \to 0}\frac{p(x+h)-p(x)}{h} \\ &= \lim_{h \to 0}\frac{4(x^2+2xh+h^2)-4x-4h-4x^2+4x}{h} \\ &= \lim_{h \to 0}\frac{4x^2+8xh+4h^2-4h-4x^2}{h} \\ &= \lim_{h \to 0}\frac{h(8x+4h-4)}{h} \\ &= \lim_{h \to 0}(8x+4h-4) \\ p'(x) &= 8x-4 \end{align*}$$

$$\begin{align*} k(x) &= 10x^3 \\ k(x+h) &= 10(x+h)^3 \\ k'(x) &= \lim_{h \to 0}\frac{k(x+h)-k(x)}{h} \\ &= \lim_{h \to 0}\frac{10(x+h)^3-10x^3}{h} \\ &= \lim_{h \to 0}\frac{10(x^3+3x^2h+3xh^2+h^3)-10x^3}{h} \\ &= \lim_{h \to 0}\frac{10x^3+30x^2h+30xh^2+10h^3-10x^3}{h} \\ &= \lim_{h \to 0}\frac{30x^2h+30xh^2+10h^3}{h} \\ &= \lim_{h \to 0}\frac{h(30x^2+30xh+10h^2)}{h} \\ &= \lim_{h \to 0}(30x^2+30xh+10h^2) \\ k'(x) &= 30x^2 \end{align*}$$

$$\begin{align*} f(x) &= 3x^2-5x+2 \\ f(x+h) &= 3(x+h)^2-5(x+h)+2 \\ f'(x) &= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0}\frac{3(x^2+2xh+h^2)-5x-5h+2-3x^2+5x-2}{h} \\ &= \lim_{h \to 0}\frac{3x^2+6xh+3h^2-5h-3x^2}{h} \\ &= \lim_{h \to 0}\frac{6xh+3h^2-5h}{h} \\ &= \lim_{h \to 0}\frac{h(6x+3h-5)}{h} \\ &= \lim_{h \to 0}(6x+3h-5) \\ f'(x) &= 6x-5 \end{align*}$$

Now find $f'(2)$:

$$\begin{align*} f'(2) &= 6(2)-5 \\ &= 12-5 \\ &= 7 \end{align*}$$

Interpretation: The derivative of $f(x)$ at $x = 2$ is $7$. This means the gradient of the function $f$ at $x = 2$ is equal to $7$, and the gradient of the tangent to $f(x)$ at $x = 2$ is equal to $7$.