Differentiation 4: Differentiating expressions of the form kx^n with respect to x

Overview:
  1. Gradient of a line passing through two points
  2. Limits
  3. Differentiation from first principles
  4. Differentiating expressions of the form \(kx^n\) with respect to \(x\)
  5. The gradient at a point on a curve
  6. 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 \(k\) and \(n\) 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.

<