Alternative Notation

While the function notation \( f'(x) \) is commonly used, there is an alternative notation for the derivative that explicitly shows the variables involved. If we have a relationship \( y = f(x) \), the derivative is often written as:

$$ \frac{dy}{dx} $$

This notation, introduced by Leibniz, reminds us that the gradient is derived from the “difference in \( y \)” divided by the “difference in \( x \)” (\( \Delta y / \Delta x \)) as the differences become infinitely small.

Worked Example

The tangent to the curve \( y = 2x^2 + 3x – 4 \) at the point \( A \) has a gradient of 11. Find the coordinates of \( A \).

Solution

First, we find the derivative function to determine the gradient formula.

$$ \frac{dy}{dx} = 4x + 3 $$

We are told that the gradient at point \( A \) is 11. Therefore, we equate the derivative to 11.

$$ 4x + 3 = 11 $$

$$ 4x = 8 \rightarrow x = 2 $$

Now, substitute \( x = 2 \) into the original equation to find the corresponding \( y \)-coordinate.

$$ y = 2(2)^2 + 3(2) – 4 $$

$$ y = 8 + 6 – 4 = 10 $$

So, the coordinates of \( A \) are \( (2, 10) \).

Equations of Tangents and Normals

A tangent is a straight line that touches the curve at a point and has the same gradient as the curve at that point.

A normal is a straight line that passes through the point of contact and is perpendicular to the tangent.

Worked Example

Find the equation of the tangent to the curve \( y = 2x^2 + 4\sqrt{x} \) at the point where \( x = 4 \).

Solution

Step 1: Find the derivative.

Rewrite the term \( 4\sqrt{x} \) as \( 4x^{\frac{1}{2}} \) to apply the power rule.

$$ y = 2x^2 + 4x^{\frac{1}{2}} $$

$$ \frac{dy}{dx} = 4x + 4(\frac{1}{2})x^{-\frac{1}{2}} = 4x + \frac{2}{\sqrt{x}} $$

Step 2: Calculate the gradient at \( x = 4 \).

$$ \text{Gradient } m = 4(4) + \frac{2}{\sqrt{4}} = 16 + \frac{2}{2} = 17 $$

Step 3: Find the \( y \)-coordinate at \( x = 4 \).

$$ y = 2(4)^2 + 4\sqrt{4} = 2(16) + 4(2) = 32 + 8 = 40 $$

The point is \( (4, 40) \).

Step 4: Write the equation of the line.

Using the point-slope form \( y – y_1 = m(x – x_1) \):

$$ y – 40 = 17(x – 4) $$

$$ y = 17x – 68 + 40 $$

$$ y = 17x – 28 $$

Worked Example

Find the equation of the normal to the curve defined by \( f(x) = 2x^3 + 3x – 2 \) at the point where \( x = 1 \).

Solution

Step 1: Find the point of contact.

$$ f(1) = 2(1)^3 + 3(1) – 2 = 2 + 3 – 2 = 3 $$

So the point is \( (1, 3) \).

Step 2: Find the gradient of the tangent.

Differentiate the function:

$$ f'(x) = 6x^2 + 3 $$

Substitute \( x = 1 \):

$$ f'(1) = 6(1)^2 + 3 = 9 $$

The gradient of the tangent is 9.

Step 3: Find the gradient of the normal.

Since the normal is perpendicular to the tangent:

$$ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{9} $$

Step 4: Write the equation.

$$ y – 3 = -\frac{1}{9}(x – 1) $$

Multiplying by 9:

$$ 9y – 27 = -(x – 1) $$

$$ 9y – 27 = -x + 1 $$

$$ x + 9y – 28 = 0 $$

Worked Example

The gradient of the normal to the graph of the function defined by \( f(x) = kx^3 – 2x + 1 \) at the point \( (1, b) \) is \( -\frac{1}{4} \).

Find the values of \( k \) and \( b \).

Solution

Step 1: Use the normal gradient to find the tangent gradient.

If the normal gradient is \( -\frac{1}{4} \), the tangent gradient must be the negative reciprocal:

$$ m_{\text{tangent}} = 4 $$

Step 2: Find \( k \) using the derivative.

$$ f'(x) = 3kx^2 – 2 $$

At \( x = 1 \), the derivative equals the tangent gradient:

$$ f'(1) = 3k(1)^2 – 2 = 4 $$

$$ 3k – 2 = 4 $$

$$ 3k = 6 \rightarrow k = 2 $$

Step 3: Find \( b \) using the original function.

Substitute \( k = 2 \) and \( x = 1 \) into \( f(x) \):

$$ f(1) = 2(1)^3 – 2(1) + 1 $$

$$ b = 2 – 2 + 1 $$

$$ b = 1 $$

Local Maximum and Minimum Points

The derivative allows us to identify turning points on a curve. At a smooth peak (maximum) or a smooth valley (minimum), the tangent is horizontal.

Worked Example

Consider the derivative function \( \frac{dy}{dx} = 1 – \frac{1}{x^2} \) for \( x \neq 0 \).

a. Find the values of \( x \) at which \( \frac{dy}{dx} = 0 \).

b. State whether these points represent a local maximum or minimum on the curve for \( y \).

Solution

a. Finding stationary points

Set the derivative to zero:

$$ 1 – \frac{1}{x^2} = 0 $$

$$ 1 = \frac{1}{x^2} $$

$$ x^2 = 1 \rightarrow x = \pm 1 $$

So, the stationary points are at \( x = 1 \) and \( x = -1 \).

b. Determining the nature of the points

We examine the sign of the gradient around the points.

For \( x = 1 \):

Test \( x = 0.5 \): \( \frac{dy}{dx} = 1 – \frac{1}{0.25} = 1 – 4 = -3 \) (Negative)

Test \( x = 2 \): \( \frac{dy}{dx} = 1 – \frac{1}{4} = 0.75 \) (Positive)

Since the gradient goes Negative \(\to\) Zero \(\to\) Positive, \( x = 1 \) corresponds to a Local Minimum.

For \( x = -1 \):

Test \( x = -2 \): \( \frac{dy}{dx} = 1 – \frac{1}{4} = 0.75 \) (Positive)

Test \( x = -0.5 \): \( \frac{dy}{dx} = 1 – 4 = -3 \) (Negative)

Since the gradient goes Positive \(\to\) Zero \(\to\) Negative, \( x = -1 \) corresponds to a Local Maximum.

Solving Problems with Constraints

Worked Example

A can of dog food contains \( 500 \text{ cm}^3 \) of food. The manufacturer, wanting to make sure that the company received maximum profits, would like to make sure that the surface area of the can was as small as possible.

Let the radius of the can be \( r \) cm and the height, \( h \) cm.

a. Find an expression for the surface area \( S \) in terms of \( r \).

b. Find \( \frac{dS}{dr} \).

c. Hence, find the dimensions of the can that will have the minimum surface area.

Solution

a. Finding the expression for Surface Area

The surface area of a cylinder is given by the sum of the curved surface area and the two circular ends:

$$ S = 2\pi rh + 2\pi r^2 $$

However, this equation has two variables, \( r \) and \( h \). We need to eliminate \( h \) using the volume constraint.

We know the volume is \( 500 \text{ cm}^3 \):

$$ V = \pi r^2 h = 500 $$

Rearranging to solve for \( h \):

$$ h = \frac{500}{\pi r^2} $$

Now, substitute this expression for \( h \) into the surface area equation:

$$ S = 2\pi r \left( \frac{500}{\pi r^2} \right) + 2\pi r^2 $$

$$ S = \frac{1000}{r} + 2\pi r^2 $$

b. Finding the derivative

First, rewrite the term \( \frac{1000}{r} \) using a negative exponent so we can apply the power rule:

$$ S = 1000r^{-1} + 2\pi r^2 $$

Now, differentiate with respect to \( r \):

$$ \frac{dS}{dr} = -1000r^{-2} + 4\pi r $$

c. Finding the minimum dimensions

A local maximum or minimum occurs when the gradient is zero.

$$ \frac{dS}{dr} = 0 $$

$$ -1000r^{-2} + 4\pi r = 0 $$

$$ \frac{-1000}{r^2} + 4\pi r = 0 $$

Multiply by \( r^2 \) (assuming \( r \neq 0 \)) or rearrange:

$$ 4\pi r = \frac{1000}{r^2} $$

$$ 4\pi r^3 = 1000 $$

$$ r^3 = \frac{1000}{4\pi} = \frac{250}{\pi} $$

$$ r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30 \text{ cm} $$

Now we find the corresponding height \( h \) using the equation from part a:

$$ h = \frac{500}{\pi (4.30)^2} \approx 8.60 \text{ cm} $$

So, the best dimensions for the can are \( r \approx 4.3 \text{ cm} \) and \( h \approx 8.6 \text{ cm} \).