By identifying some key features of a function's curve, you can gain an understanding of what the gradient function's curve looks like. This will make use of differentiation knowledge you already have.

Gradient function

Suppose you have a function $f(x)$ . Its curve is given by $y=f(x)$ . The gradient function is found by differentiating once: $\dfrac{\text dy}{\text dx}=f'(x)$ . But this is a function itself and can be sketched.

Example 1

Consider the curve given by $y=x^3-3x$ :

Maths; Differentiation I; KS5 Year 12; Sketching gradient functions

One method to sketch the gradient function curve would be to differentiate ( $\dfrac{\text dy}{\text dx}=3x^2-3$ ) and sketch the corresponding curve: $y=3x^2-3$ :

Sketching the gradient function without differentiation

Just because you can sketch the curve of a function, it does not mean you will always be able to easily sketch the curve of the gradient function. Moreover, sometimes you will only have the sketch of the function without the function itself. In this case, you cannot use differentiation to obtain the gradient function.

Instead, you will rely on key features of the function's curve to identify what the key features of the gradient function's curve are.

Features of $y=f(x)$	Features of $y=f'(x)$	Explanation
Positive gradient	Above the $x$ axis	If the gradient of $y=f(x)$ is positive, then $f'(x)>0$ and hence for these values of $x$ , the gradient function curve has positive values of $y$ .
Negative gradient	Below the $x$ axis	If the gradient of $y=f(x)$ is negative, then $f'(x)<0$ and hence for these values of $x$ , the gradient function curve has negative values of $y$ .
Minimum or maximum	Cuts the $x$ axis	A maximum or minimum is a stationary point, so the gradient is zero. In other words, $f'(x)=0$ and so on the curve $y=f'(x)$ it follows that $y=0$ for this $x$ . In particular, the gradient of the function curve is changing from positive to negative (or negative to positive), so it follows that on the gradient function curve, you go from above the $x$ axis to below (or from below to above) i.e. cutting the $x$ axis.
Point of inflection	Touches the $x$ axis	Since a point of inflection is also a stationary point, it follows that for this $x$ , the gradient function curve is at the $x$ axis. However, the gradient of the function goes from positive to zero to positive (or negative to zero to negative), so it follows that the gradient function curve only touches the $x$ axis without passing through it.
Vertical asymptote	Vertical asymptote	If a function's curve has a vertical asymptote, it follows that it's gradient is tending to positive or negative infinity. Thus, the gradient function's curve will also have the same asymptote.
Horizontal asymptote	Horizontal asymptote at $y=0$	A horizontal asymptote on $y=f(x)$ means the gradient is tending to zero, thus the curve $y=f'(x)$ must tend to the $x$ axis.

Example 2

Given the following curve for some function, sketch a curve of its gradient function.

Identify the (rough) $x$ -coordinates of the turning points (maxima and minima). These will be the same $x$ -coordinates as the $x$ -intercepts on the gradient function curve. The left turning point (the local minimum) has an $x$ -coordinate of about $-1.5$ . The right turning point (the local maximum) has an $x$ -coordinate of about $0.9$ .

Next notice that between the turning points, the gradient is positive, so on the gradient function curve, the curve will be above the $x$ -axis between the $x$ -intercepts. The gradient is negative to the left of the minimum and also to the right of the maximum. Thus, on the gradient function curve, the curve will be below the $x$ -axis to the left of the left $x$ -intercept and also to the right of the right $x$ -intercept.

Sketching this gives:

Putting them together on the same grid makes their relationship clearer:

Dotted lines are given to indicate that the turning points on the function curve correspond to $x$ -intercepts on the gradient function curve.

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Unit 1

Finding the gradient of a straight line

Unit 2

Finding the gradient of a curve

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Unit 3