Review question

# If we increase $x$ by a small amount, what happens to $y$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9942

## Solution

Given that $y = 4x^3 - 7x^2$, find the value of $dy/dx$ at the point $(2,4)$.

We know that the derivative of $x^n$ is $nx^{n-1}$. Using this, we have that $\frac{dy}{dx} = y' = 4 \times 3x^2 - 7 \times 2x = 12x^2 - 14x.$ So we have that the value of $\dfrac{dy}{dx}$ at the point $(2,4)$ is $y'(2) = 12 \times 2^2 - 14 \times 2 = 48 - 28 = 20.$

Hence find the approximate increase in $x$ which will cause $y$ to increase from $4$ to $4.05$.

Consider the following graph of a curve $y = f(x)$.

The slope of the tangent is close to the slope of the chord; that is, to use the notation in the figure, $\frac{\delta y}{\delta x} \approx \frac{dy}{dx} \quad\text{when \delta x is small.}$ So if the $x$-coordinate of the leftmost point in this figure is denoted by $x_0$, then $\begin{equation}\label{eq:approximation-to-derivative} \frac{f(x_0+\delta x) - f(x_0)}{\delta x} \approx f'(x_0) \quad\text{when \delta x is small}. \end{equation}$

In fact, the definition of $f'(x_0)$ is the value towards which $\frac{f(x_0+\delta x) - f(x_0)}{\delta x}$ tends as $\delta x$ tends towards zero.

In our case:

• $f(x) = 4x^3 - 7x^2$,
• $x_0 = 2$,
• $f(x_0) = 4$, and
• $f'(x_0) = 20$.

If we choose $\delta x$ so that $f(x_0 + \delta x) = 4.05$, then, from $\eqref{eq:approximation-to-derivative}$, we would have $\frac{4.05 - 4}{\delta x} = \frac{f(x_0+\delta x) - f(x_0)}{\delta x} \approx f'(x_0) = 20,$ and so, by rearranging, $\delta x \approx \frac{4.05 - 4}{20} = \frac{1/20}{20} = \frac{1}{400} = 0.0025.$

If we use our calculators, we find that $f(2.0025) - f(2) = 0.05010\ldots,$ and so our approximation is a good one!