Introduction

This is the second part of first series — Derivatives, focused on the concept of gradient functions.

Note: the content introduced in this series is mostly based on the IGCSE Further Math textbook.

Quick Access

• Part 1.1 - Basic Differentiation Rules
• Part 1.2 - Gradient Functions
• Part 1.3 - Rate of Change
• Part 1.4 - Using the Rate of Change
• Part 2.1 - Definite & Indefinite Integrals
• Part 2.2 - Area Under the Curve

Gradient Functions

Now that you have some basic knowledge of the different rules and formulae used to differentiate functions. Let's talk a bit more about the graphical interpretations.

The first derivative of a function is always the gradient function of the curve/line.

For example, for a linear function $y = mx + b$, its derivative is given by $\frac{dy}{dx} = m$, which means that for any linear function, its derivative is always its gradient.

Now, when it comes to a non-linear function, for any arbitrary point $P$ on the curve, it is said to have its own gradient, and its gradient is that of the tangent to the curve at P.

The green line is the tangent to the blue curve at P. Try to drag the point $P$, and you'll see that the gradient of the tangent varies as $P$ moves along the curve.

Thus the gradient of a curve changes, unlike a linear equation where the gradient is constant no matter what $x$ is.

Take point $P(x, f(x))$, how do we find the gradient at $P$ then?

As discussed above, we can formulate a gradient function for any line/curve by taking the derivative of the original function.

In this example, the curve shown is a quadratic function, so let $f(x) = ax^2 + bx + c$. Taking its derivative would give $f'(x) = 2ax + b$, and the tangent's gradient at $P(x, f(x))$ is simply equal to $f'(x)$.

Calculating the Equation of The Tangent

Now that we have the gradient of the tangent, you might wonder, can we form an equation for the straight line?

We can use the formula for the equation of a straight line $y - y_1 = m(x - x_1)$, and in this case, $x_1$ and $y_1$ would be the coordinates of the point $P$.

The equation of the tangent to a curve at $P(a, f(a))$ is:

The Normal

The normal is a straight line that crosses the tangent at 90 degrees (perpendicular) through the point where the tangent meets the curve.

Therefore, the normal's gradient is the negative reciprocal to that of the tangent.

Hence we can conclude that the equation of the normal can be calculated as follows:

Example

The equation of a curve is given by $f(x) = x^3 - 3x^2 + 2x - 1$, where $y = f(x)$, find the equations of the tangent and normal at the point (3, 5).

1. Find the gradient function by taking the function's derivative and calculating $f'(3)$

2. Find the equation of the tangent

3. Find the equation of the normal

Stationary Points

Stationary points are points on a curve such that their tangent has a gradient of 0 (i.e., a horizontal line). They are said to be stationary because, at these points, the function is neither decreasing nor increasing.

A function can have multiple stationary points, or just one. For example, a quadratic curve ($y = ax^2 + bx + c$) would have only one stationary point.

There are three types of stationary points:

• Maximum
• Minimum
• Neither

For example, the point labeled in the image is a minimum stationary point.

A stationary point is a turning point if it is where the curve changes from increasing to decreasing, or the other way around.

In the image above, the labeled point is a minimum turning point, because from there, the function becomes increasing.

A polynomial function of degree $n$ would have at most $n - 1$ turning points. E.g., a quadratic function (degree 2) has at most $2 -1 = 1$ turning points.

Finding the Stationary Points

As said above, a stationary point has a gradient of 0. So to calculate the coordinates of the stationary points of a function set $f'(x) = 0$ and find the solutions of x

These values of x would be the x-coordinates of the stationary points, and to find the corresponding y-coordinates, simply calculate $f(x)$ for each x-coordinate.

Example

The function of the curve above is given by $f\left(x\right)\ =\ -x^{2\ }+3x+2$; find the coordinates of the stationary point P.

1. Find the derivative of $f(x)$

2. Solve for $x$ when $f'(x) = 0$

3. Substitute the $x$ value back into $f(x)$

Conclusion

Stationary point $P = (1\frac{1}{2}, 4\frac{1}{4})$.

The Second Derivative

Now, what if the curve we are dealing with has more than one stationary point? How do we classify them into these three categories?

The second derivative of a function can be used to check if a stationary point is the maximum, minimum, or neither.

Key Point

If $f'(x) = 0$, i.e. for a stationary point:

• If $f''(x) < 0$, the point is a maximum point.
• Else if $f''(x) > 0$, the point is a minimum point.
• Otherwise, the point is neither a maximum nor a minimum.

Example

The curve is given by the equation $f\left(x\right)\ =\ x\left(x+2\right)\left(x-2\right)$. Find the stationary points of the function and determine which of them are maximum/minimum.

To find the coordinates of the two stationary points:

1. Expand

2. Differentiate

3. Set $f'(x) = 0$ and solve for x

4. Find the corresponding y coordinates

So our coordinates would be $\left(\pm\frac{2\sqrt{3}}{3}, \mp\frac{16\sqrt{3}}{9}\right)$

Now, determine which is maximum or minimum:

1. Find the second derivative of the function

2. So for any positive value of $x$, the point is a minimum, and hence for any negative value of $x$, the point is a maximum.

Conclusion

• $x_1 = \frac{2\sqrt{3}}{3} > 0$, so the stationary point with the x-coordinate $x_1$ is a minimum.

• $x_2 = -\frac{2\sqrt{3}}{3} < 0$, so the stationary point with the x-coordinate $x_2$ is a maximum.

Conclusion

Subsequent sections are accessible through this section at the top of the article.

Thanks for reading.