The notation dy/dx for the derivative
Small changes Δx and Δy, Leibniz notation dy/dx, operator notation d/dx, and second and higher derivatives.
Small changes and dy/dx
When $x$ changes by a small amount $\Delta x$, $y$ changes by $\Delta y$. The ratio $\dfrac{\Delta y}{\Delta x}$ is the gradient of a secant. As $\Delta x \to 0$ it becomes the gradient of the tangent:
$\dfrac{dy}{dx}$ and $f'(x)$ mean the same thing. The operator $\dfrac{d}{dx}$ means "differentiate with respect to $x$", as in $\dfrac{d}{dx}(x^3) = 3x^2$. Other letters work too: if $s = 3t^2$, then $\dfrac{ds}{dt} = 6t$.
Second and higher derivatives
Differentiating again gives the second derivative, written $f''(x)$ or $\dfrac{d^2y}{dx^2}$. It measures how fast the gradient itself is changing.
Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ for $y = x^4 - 3x^2$.
Find the third derivative of $x^5$.
$\dfrac{dy}{dx}$ means:
It is the same as $f'(x)$.
For $y = x^4 - 3x^2$, find $\dfrac{d^2y}{dx^2}$ at $x = 1$.
$12 - 6 = 6$.
The third derivative of $x^5$ is:
$5x^4 \to 20x^3 \to 60x^2$.
Find $\dfrac{d}{dx}(5x^2 - 3x)$ at $x = 2$.
$10x - 3 = 17$.
$\dfrac{\Delta y}{\Delta x}$ is the gradient of:
It uses two separate points.
If $f(x) = 2x^3$, find $f''(-1)$.
$f''(x) = 12x$.
If $s = 3t^2 - t^3$, find $\dfrac{ds}{dt}$ at $t = 1$.
$6t - 3t^2 = 3$.
For which function is $f''(x) = 0$ for every $x$?
A linear function has a constant gradient.
For $y = x^3 - 6x^2$, find $x$ where $\dfrac{d^2y}{dx^2} = 0$.
$6x - 12 = 0$.
New numbers every time. Keep going until you get five in a row.
Coaching Centre