11. Differentiation

iGCSE (2021 Edition)

Lesson

For functions of the type $f(x)=x^n$`f`(`x`)=`x``n`:

$f(x)$f(x) |
$f'(x)$f′(x) |
---|---|

$x$x |
$1$1 |

$x^2$x2 |
$2x$2x |

$x^3$x3 |
$3x^2$3x2 |

$x^4$x4 |
$4x^3$4x3 |

$x^5$x5 |
$5x^4$5x4 |

Noticing the pattern in the table, we could make the conjecture that the derivative of $f(x)=x^n$`f`(`x`)=`x``n` is $f'(x)=nx^{n-1}$`f`′(`x`)=`n``x``n`−1. This is, in fact, the case and we call this a rule for differentiating powers of x. This applies to not only positive integer values of $n$`n` but any real number value of $n$`n`.

Differentiation of $f(x)=x^n$`f`(`x`)=`x``n`:

For a function of the form $f(x)=x^n$`f`(`x`)=`x``n` the derivative is:

$f'(x)=nx^{n-1}$`f`′(`x`)=`n``x``n`−1

Find the derivative of $f(x)=4x^3$`f`(`x`)=4`x`3.

**Think:** When the pronumeral in the function has a coefficient, in this case, $4$4, you just find the derivative of the part of the function with the pronumeral and multiply it by the coefficient.

**Do:**

$f(x)$f(x) |
$=$= | $4x^3$4x3 |

$f'(x)$f′(x) |
$=$= | $4\times\ 3x^2$4× 3x2 |

$f'(x)$f′(x) |
$=$= | $12x^2$12x2 |

Let's look at a harder example:

Find the derivative of $f(x)=\frac{1}{2x^2}$`f`(`x`)=12`x`2.

**Think:** Recall from index laws that $\frac{1}{x^n}=x^{-n}$1`x``n`=`x`−`n`, and as mentioned above our rule for differentiation works for all real values of $n$`n`. It also useful to separate the coefficient from the $x$`x` term to make the process easier and less prone to error.

**Do:**

$f(x)$f(x) |
$=$= | $\frac{1}{2x^2}$12x2 |

$f(x)$f(x) |
$=$= | $\frac{1}{2}\times\frac{1}{x^2}$12×1x2 |

$f(x)$f(x) |
$=$= | $\frac{1}{2}x^{-2}$12x−2 |

$f'(x)$f′(x) |
$=$= | $\frac{1}{2}\times-2x^{-3}$12×−2x−3 |

$f'(x)$f′(x) |
$=$= | $-x^{-3}$−x−3 |

If there are multiple terms in a function we just apply the rule for differentiation that we have learnt to each term and add or subtract them. The rule shown in the above example can be written as:

Composite functions

For $f(x)=g(x)+h(x)$`f`(`x`)=`g`(`x`)+`h`(`x`):

$f'(x)=g'(x)+h'(x)$`f`′(`x`)=`g`′(`x`)+`h`′(`x`)

Find the gradient of $f(x)=x^2-3x$`f`(`x`)=`x`2−3`x` at the point $(4,14)$(4,14)

**Think:** to find the derivative of this function, we consider and differentiate each term individually and then add them together.

**Do:**

$f(x)$f(x) |
$=$= | $x^2-3x$x2−3x |

$f'(x)$f′(x) |
$=$= | $2x-3$2x−3 |

To find the gradient at the point $(4,14)$(4,14) just substitute $x=4$`x`=4 into $f'(x)$`f`′(`x`):

$f'(x)$f′(x) |
$=$= | $2\times4-3$2×4−3 |

$f'(4)$f′(4) |
$=$= | $5$5 |

Find the gradient of $f\left(x\right)=x^4$`f`(`x`)=`x`4 at $x=2$`x`=2.

Denote this gradient by $f'\left(2\right)$`f`′(2).

Consider the function $y=\frac{8x^9-4x^8+6x^7+9}{2x^2}$`y`=8`x`9−4`x`8+6`x`7+92`x`2.

Rewrite the function so that each term is a power of $x$

`x`.Find the derivative of the function.

Find the gradient of $f\left(x\right)=x^5-3x^4$`f`(`x`)=`x`5−3`x`4 at the point $\left(3,0\right)$(3,0).

Denote this gradient by $f'\left(3\right)$`f`′(3).

We want to differentiate $y=2x^2\left(7x+2\right)$`y`=2`x`2(7`x`+2).

Rewrite the equation in expanded form.

Differentiate the rewritten equation.

Find the derivative of $y=x^{-5}+7$`y`=`x`−5+7.

Give your answer with positive indices.

Use the notations f'(x), f''(x), dy/dx, d^2y/dx^2 [=d/dx(dy/dx)].