Further Differentiation
Further Differentiation
1. Increasing and Decreasing Functions
A function y = f(x) is:
- Increasing: If
dy/dx > 0throughout the interval. - Decreasing: If
dy/dx < 0throughout the interval.
2. Stationary Points
Stationary points, also known as turning points, occur when:
dy/dx = 0
3. First Derivative Test for Maximum and Minimum Points
- At a maximum point:
dy/dx = 0- The gradient is positive to the left and negative to the right of the point.
- At a minimum point:
dy/dx = 0- The gradient is negative to the left and positive to the right of the point.
4. Second Derivative Test for Maximum and Minimum Points
- If
dy/dx = 0andd²y/dx² < 0, the point is a maximum. - If
dy/dx = 0andd²y/dx² > 0, the point is a minimum. - If
dy/dx = 0andd²y/dx² = 0, use the first derivative test to determine the nature of the stationary point.
5. Connected Rates of Change
If two variables, x and y, both vary with a third variable, say time (t):
- The chain rule applies:
- You can also use:
Further Differentiation: Examples and Solutions
Example 1
The variables \(x\), \(y\), and \(z\) are such that \(z = 3x + \frac{1200}{x}\). Show that:
- \(\frac{dz}{dx} = 3 - \frac{1200}{x^2}\).
- Find the stationary value of \(z\) and determine its nature.
Solution:
\(z = 3x + \frac{1200}{x}\)
\(\frac{dz}{dx} = 3 - \frac{1200}{x^2}\).
To find the stationary point, set \(\frac{dz}{dx} = 0\):
\(3 - \frac{1200}{x^2} = 0 \implies x^2 = 400 \implies x = 20\) (positive value).
Second derivative test:
\(\frac{d^2z}{dx^2} = \frac{2400}{x^3}\), which is positive for \(x > 0\), so \(z\) has a minimum at \(x = 20\).
Example 2
A curve has the equation \(y = \frac{8}{x} + 2x\). Find:
- \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\).
- The coordinates of the stationary points and their nature.
Solution:
\(y = \frac{8}{x} + 2x\)
\(\frac{dy}{dx} = -\frac{8}{x^2} + 2\)
\(\frac{d^2y}{dx^2} = \frac{16}{x^3}\)
Stationary points occur when \(\frac{dy}{dx} = 0\):
\(-\frac{8}{x^2} + 2 = 0 \implies x^2 = 4 \implies x = \pm 2\).
- For \(x = 2\): \(\frac{d^2y}{dx^2} > 0\), so it is a minimum.
- For \(x = -2\): \(\frac{d^2y}{dx^2} > 0\), so it is also a minimum.
Example 3
The volume of a sphere is increasing at a constant rate of \(40 \, \text{cm}^3/\text{s}\). Find the rate of increase of the radius when the radius is \(15\, \text{cm}\). Use:
\(V = \frac{4}{3}\pi r^3\).
Solution:
\(V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\).
Given that \(\frac{dV}{dt} = 40\, \text{cm}^3/\text{s}\) and \(r = 15\, \text{cm}\):
\(40 = 4\pi (15)^2 \cdot \frac{dr}{dt}\).
Solve for \(\frac{dr}{dt}\):
\(\frac{dr}{dt} = \frac{40}{4\pi (15)^2} = \frac{40}{900\pi} = \frac{4}{90\pi}\, \text{cm/s}\).
