The derivative function and the rules of differentiation

The derivative can be worked out at every point at once: write $\displaystyle x$ instead of a number $\displaystyle x_0$. This gives a new function, the derivative function $\displaystyle f'(x)$, which gives every $\displaystyle x$ the slope of the tangent there.

The example below finds the derivative of $\displaystyle {f(x)}={x^2}$ by the definition:

• As $\displaystyle h\to0$, $\displaystyle 2x+h$ tends to $\displaystyle 2x$, so $\displaystyle {f'(x)}={2x}$.
• Now every derivative at a point is a substitution: $\displaystyle {f'(1)}={2}$, $\displaystyle {f'(3)}={6}$, $\displaystyle {f'(-2)}={-4}$.

The same way, by the definition:

• $\displaystyle {(x^3)'}={3x^2}$ (Exercise 1);
• $\displaystyle {(x)'}={1}$: the graph of $\displaystyle {y}={x}$ is a line of slope 1;
• $\displaystyle {(c)'}={0}$ for every constant number $\displaystyle c$: the graph of $\displaystyle {y}={c}$ is a horizontal line, of slope 0.

All of these are cases of one rule – the power comes down in front, and the exponent goes down by 1:

For example $\displaystyle {(x^5)'}={5x^4}$, and $\displaystyle {(x^{10})'}={10x^9}$.

The proof below shows, by the definition, that the derivative of $\displaystyle c\cdot f(x)$ is $\displaystyle c\cdot f'(x)$.

• $\displaystyle {g(x)}={c\cdot f(x)}$, so the slope of a secant of $\displaystyle g$ is $\displaystyle {\frac{c\cdot f(x+h)-c\cdot f(x)}{h}}={c\cdot\frac{f(x+h)-f(x)}{h}}$.
• As $\displaystyle h\to0$, the fraction $\displaystyle \frac{f(x+h)-f(x)}{h}$ tends to $\displaystyle f'(x)$.
• So $\displaystyle c$ times the fraction tends to $\displaystyle c\cdot f'(x)$, that is $\displaystyle {g'(x)}={c\cdot f'(x)}$.

On the graph: $\displaystyle {y}={c\cdot f(x)}$ is a vertical stretch of the graph of $\displaystyle f$ by $\displaystyle c$, and all the slopes are stretched with it by $\displaystyle c$. On the board below are $\displaystyle {f(x)}={x^2}$ and $\displaystyle {g(x)}={3x^2}$, and their tangents at $\displaystyle {x}={1}$:

• The slope of the tangent to $\displaystyle f$ at $\displaystyle {x}={1}$ is $\displaystyle {f'(1)}={2}$.
• The slope of the tangent to $\displaystyle g$ there is $\displaystyle {g'(1)}={3\cdot2}={6}$ – 3 times as much.

The proof below shows, by the definition, that the derivative of $\displaystyle f(x)+g(x)$ is $\displaystyle f'(x)+g'(x)$.

• $\displaystyle {k(x)}={f(x)+g(x)}$, and the slope of a secant of $\displaystyle k$ is $\displaystyle \frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}$.
• Rearrange the numerator and split the fraction: $\displaystyle \frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}$.
• As $\displaystyle h\to0$, the first fraction tends to $\displaystyle f'(x)$ and the second to $\displaystyle g'(x)$.
• So $\displaystyle {k'(x)}={f'(x)+g'(x)}$.

The derivative of a difference is proved the same way (Exercise 4).

Together the three rules differentiate any polynomial, term by term. The example below differentiates $\displaystyle {f(x)}={4x^3-5x^2+7x-2}$:

And the slope of the tangent at $\displaystyle {x}={1}$: $\displaystyle {f'(1)}={12-10+7}={9}$.

The rules know nothing of a product: expand $\displaystyle (x-3)(x+2)$ into a polynomial first, $\displaystyle x^2-x-6$, and only then differentiate.

When $\displaystyle s(t)$ is the distance (or the height) against time, its derivative is the instantaneous speed, and the derivative of the speed is the acceleration – the rate of change of the speed:

In the example below a ball is kicked from the ground straight up at 30 metres per second. By Newton’s laws of motion, its height after $\displaystyle t$ seconds is $\displaystyle {y(t)}={30t-5t^2}$ metres (the number 5 is half the acceleration of gravity, which is about 10 metres per second per second).

• The speed: $\displaystyle {v(t)}={y'(t)}={30-10t}$. At $\displaystyle {t}={0}$ it is 30 – the speed the ball was kicked at.
• The acceleration: $\displaystyle {a(t)}={v'(t)}={-10}$. The speed goes down by 10 metres per second every second – gravity slows the ball down.
• At $\displaystyle {t}={3}$ the speed is 0: the ball stops rising and starts to fall. Its height then is $\displaystyle {y(3)}={90-45}={45}$ metres.
• $\displaystyle {y(t)}={0}$ when $\displaystyle {t(30-5t)}={0}$, that is at $\displaystyle {t}={6}$: the ball lands.

We find the derivative of $\displaystyle {f(x)}={x^3}$ by the definition.

Simplify the slope of the secant $\displaystyle \frac{(x+h)^3-x^3}{h}$.
Use $\displaystyle {(x+h)^3}={x^3+3x^2h+3xh^2+h^3}$.

What does your expression tend to as $\displaystyle h\to0$?
Write the derivative of $\displaystyle x^3$.

Differentiate.

The function is $\displaystyle {f(x)}={x^3-6x}$.

Work out the slope of the tangent to the graph of $\displaystyle f$ at the point where $\displaystyle {x}={2}$.

At which points on the graph is the slope of the tangent 6?
Write an equation and solve it.

Prove by the definition: $\displaystyle {(f(x)-g(x))'}={f'(x)-g'(x)}$.

A ball is kicked from the ground straight up.
Its height after $\displaystyle t$ seconds is $\displaystyle {y(t)}={40t-5t^2}$ metres.

Find the ball’s speed $\displaystyle v(t)$.

When does the ball stop rising?
Write an equation and solve it.

How high does the ball go?

Find the ball’s acceleration: differentiate the speed you found in part a.

When does the ball land?
What is its speed at the moment it lands, and what does its sign mean?