Coaching Centre
The derivative as a limit
The tangent as the limit of secants, differentiating from first principles, finding points with a given gradient, and where a function is not differentiable.
The tangent as a limit of secants
Take a point $P$ at $x$ and a nearby point $Q$ at $x + h$. The secant $PQ$ has gradient $\dfrac{f(x + h) - f(x)}{h}$. As $h \to 0$, $Q$ slides towards $P$ and the secant becomes the tangent.
Orange: y = x². Blue dashed: the secant. Drag h towards 0 and watch the secant gradient approach 2a.
With $a = 1$, what is the secant gradient when $h = 0.01$?
Differentiate $f(x) = x^2$ from first principles.
For $f(x) = x^2 - 3x$, the derivative is $f'(x) = 2x - 3$. Where is the gradient $5$?
Differentiable functions
A function is differentiable at a point if the curve is smooth and continuous there, so it has one clear tangent. It is not differentiable at a break, at a sharp corner (such as $y = |x|$ at $x = 0$), or where the tangent is vertical.
Find the gradient of the secant on $y = x^2$ from $x = 2$ to $x = 2.1$.
$2a + h = 4 + 0.1 = 4.1$.
For $f(x) = x^2$, $\dfrac{f(x + h) - f(x)}{h}$ simplifies to:
$\dfrac{2xh + h^2}{h} = 2x + h$.
Using first principles, find $f'(3)$ for $f(x) = x^2$.
$f'(x) = 2x$, so $f'(3) = 6$.
From first principles, $f(x) = 3x^2$ gives:
The quotient is $6x + 3h$, which tends to $6x$.
$f(x) = x^2 - 3x$ has $f'(x) = 2x - 3$. Find $x$ where the gradient is $5$.
$2x - 3 = 5$.
$y = |x|$ is not differentiable at $x = 0$ because:
The gradient jumps from $-1$ to $1$.
For $f(x) = x^3$, $\dfrac{f(1 + h) - f(1)}{h} = 3 + 3h + h^2$. Find $f'(1)$.
Let $h \to 0$.
$f(x) = x^2 + 5x$ has $f'(x) = 2x + 5$. Find the gradient at $x = -1$.
$-2 + 5 = 3$.
Substituting $h = 0$ straight into the difference quotient gives:
That is why you simplify first.
New numbers every time. Keep going until you get five in a row.