- Gradient of a line passing through two points
- Limits
- Differentiation from first principles
**Differentiating expressions of the form \(kx^n\) with respect to \(x\)**- The gradient at a point on a curve
- Tangents and normals

# Part 4: Differentiating expressions of the form \(kx^n\) with respect to \(x\)

In part 3, you hopefully found from first principles that if \(y=x^3\), then \(\dfrac{\text{d}y}{\text{d}x}=3x^2\), and that if \(y=5x^2\), then \(\dfrac{\text{d}y}{\text{d}x}=10x\).

#### Differentiating from first principles is time-consuming, but fortunately we can save a lot of time by noting the following

If \(y=x^n\), then \(\dfrac{\text{d}y}{\text{d}x}=nx^{n-1}\), where \(n\) is a constant.

Note that \(n\) need not be an integer.

Let’s see a few examples:

- If \(y=x^2\), then \(\dfrac{\text{d}y}{\text{d}x}=2x^{1}=2x\). Note that this agrees with what we saw in part 3.
- \(\dfrac{\text{d}}{\text{d}x}\left(x^{4}\right)=4x^3\).
- If \(y=x\) i.e. \(y=x^1\), then \(\dfrac{\text{d}y}{\text{d}x}=1x^{0}=1\). Note that this fits in with what you should know about the gradient of the line \(y=x\).
- If \(y=\dfrac{1}{x^3}\) i.e. \(y=x^{-3}\), then \(\dfrac{\text{d}y}{\text{d}x}=-3x^{-4} = -\dfrac{3}{x^4}\).
- If \(\text{f}(x)=\sqrt{x}\) i.e. \(y=x^{\frac{1}{2}}\), then \(\text{f’}(x)=\dfrac{1}{2}x^{-\frac{1}{2}}=\dfrac{1}{2\sqrt{x}}\).

#### Working with coefficients

By noting the following fact, we can also easily work with expressions with coefficients other than 1

\(\dfrac{\text{d}}{\text{d}x}\left(k \text{f}(x)\right) = k\dfrac{\text{d}}{\text{d}x}\left(\text{f}(x)\right)\)

This means that if \(y=kx^n\), then \(\dfrac{\text{d}y}{\text{d}x}=nkx^{n-1}\), where \(k\) and \(n\) are constants.

Again, note that \(k\) and \(n\) need not be integers.

Let’s see a few examples:

- If \(\text{f}(x)=5x^7\), then \(\text{f’}(x)=35x^6\).
- If \(y=7x\) i.e. \(y=7x^1\), then \(\dfrac{\text{d}y}{\text{d}x}=7x^{0}=7\). Note that this fits in with what you should know about the gradient of the line \(y=7x\).
- If \(y=8\) i.e. \(y=8x^0\), then \(\dfrac{\text{d}y}{\text{d}x}=0\).
- \(\dfrac{\text{d}}{\text{d}x}\left(7x^{-\frac{3}{2}}\right)=-\dfrac{21}{2}x^{-\frac{5}{2}}\).

Note: if you are not comfortable with fractional and negative indices and how these relate to roots and reciprocals, make sure you revise those topics before continuing. It will be **much** better for you in the long run if you build on a solid base rather than build on shaky foundations.

#### A further useful fact: we can differentiate term-by-term:

\(\dfrac{\text{d}}{\text{d}x}\left(\text{f}(x) + \text{g}(x)\right) = \dfrac{\text{d}}{\text{d}x}\left(\text{f}(x)\right) + \dfrac{\text{d}}{\text{d}x}\left(\text{g}(x)\right)\)

Let’s see a few examples of how we can use this:

- If \(y=x^8+10x^3\), then \(\dfrac{\text{d}y}{\text{d}x}=8x^7+30x^2\).
- If \(y=x^2+9x+7\), then \(\dfrac{\text{d}y}{\text{d}x}=2x+9\).
- If \(y=x^{19} + 7x + 3x^{-3}\), then \(\dfrac{\text{d}y}{\text{d}x}=19x^{18}+7-9x^{-4} = 19x^{18}+7-\dfrac{9}{x^4}\).
- If \(\text{f}(x)=7x^{-\frac{3}{2}}+87\), then \(\text{f’}(x)=-\dfrac{21}{2}x^{-\frac{5}{2}}\).

### Practice questions

Use the applet below to practise more questions and check your answers. Use the slider to reveal the solutions.