Gradients, tangents and derivatives

Gradients and tangents help us understand how functions change. We use derivatives to find these. Learn how these concepts work together to analyse and interpret the behaviour of functions.

Tangent at a point along a curve

A tangent is a line that touches a curve at only one point. Where that point sits along the function curve determines the slope (i.e. the gradient) of the tangent to that point.

We can find the gradient of a tangent at any point on a curve \(f(x)\) using the derivative of the function, \(f'(x)\). If we plug an \(x\) value into the derivative function, we will get the slope of the tangent at that point \(x\).

Here, the derivative functions \(f'(x)\) have been given to you. To learn how to find derivatives, go to Differentiation from first principles.

Gradient of a tangent

You should already be familiar with how to find the gradient of a straight line. If not, you can go to Linear functions for a recap. For straight lines, the gradient is the same for the entire function.

When we are concerned with curves, the gradient can differ depending on where we are on the curve. Let's consider the curve \(f(x)\) with point \(P\).

We can approximate the tangent at point \(P(x,f(x))\) by using another point \(Q\) on the curve, which has coordinate \(Q(x+h,f(x+h))\). The approximate gradient would be:
\[\textrm{Gradient of }PQ = \frac{f(x+h)-f(x)}{h}\]

As \(h\) decreases and \(Q\) gets closer to \(P\), the approximate of the gradient becomes more accurate. In other words, the gradient becomes more accurate as \(h\) approaches \(0\). So, we can determine a more accurate gradient by finding the limit:

\[f'(x) = \lim_{h\rightarrow0}\left(\frac{f(x+h)-f(x)}{h}\right)\quad h\neq0\]

The gradient formula for the curve \(y=f(x)\) is defined as the derivative function. Thus, the derivative function \(f'(x)\) gives the slope of the tangent to the curve \(f(x)\) at any point \(x\).

Example – finding the gradient of a tangent

The derivative of the function \(f(x)=\dfrac{3}{x}\) is \(f'(x)=-\dfrac{3}{x^{2}}\). Find the slope of the tangent to the curve at \(x=4\).

At \(x=4\):
\[\begin{align*} f'(x) & =-\frac{3}{x^{2}}\\
f'(4) & = -\frac{3}{4^{2}}\\
& = -\frac{3}{16}
\end{align*}\]

The slope of the tangent at \(x=4\) is \(-\dfrac{3}{16}\).

Exercise – finding the gradient of a tangent

1. If the derivative function for \(f(x)=x^{3}-x\) is \(f'(x)=3x^{2}-1\), find the slope of the tangent to this curve at:
1. \(x=2\)
2. \(x=0\)
3. \(x=-9\)
2. If the derivative function of \(f(x)=\sin\left(x\right)\) is \(f'(x)=\cos\left(x\right)\), find the gradient of \(f(x)=\sin\left(x\right)\) at:
1. \(x=0\)
2. \(x=\dfrac{\pi}{2}\)
3. \(x=3.5\)
3. Determine \(\lim_{h\rightarrow0}\left(\frac{(x+h)^{2}-x^{2}}{h}\right)\) and hence find the slope of the tangent to the curve \(y=x^{2}\) at:
1. \(x=2\)
2. \(x=0\)
3. \(x=-9\)

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