11. Differentiation

iGCSE (2021 Edition)

Lesson

So far this chapter you have learnt that, given a function $f(x)$`f`(`x`), it is possible to find a derivative function $f'(x)$`f`′(`x`) which we can use to find the gradient of the tangent at any point on the original function. Remember, the tangent is defined as a straight line or plane that touches a curve or curved surface at a single point.

In this lesson we will look at the definition of the angle of inclination of a line and how the derivative allows us to determine the equations of tangents and normals to a function given a single point.

A normal at a point is a line that is perpendicular to the tangent at the same point on the curve.

Since the normal and tangent at a point are perpendicular to each other their gradients are related by:

$m_1\ m_2=-1$`m`1 `m`2=−1

Or:

$m_2=-\frac{1}{m_1}$`m`2=−1`m`1

Where $m_1$`m`1 and $m_2$`m`2 are the gradients of the tangent and normal. That is, the gradient of the normal is the negative reciprocal of the gradient of the tangent at the same point on a function.

Finding equations of tangents and normals requires knowing how to find the equation of a straight line. Remember that we need a point $(x_1,y_1)$(`x`1,`y`1) and the gradient $m$`m`.

Knowing these two things we can use the point gradient formula to determine the equation of a line.

Point gradient formula

$y-y_1=m(x-x_1)$`y`−`y`1=`m`(`x`−`x`1)

Find the point on the curve $y=3-4x^2$`y`=3−4`x`2 where the gradient is $16.$16.

**Think:** We will take the derivative of $y$`y` and equate it to the gradient given.

**Do: **

$\frac{dy}{dx}$dydx |
$=$= | $-8x$−8x |

Substituting $\frac{dy}{dx}=16$`d``y``d``x`=16 gives:

$-8x$−8x |
$=$= | $16$16 |

Solving for $x$`x`:

Solving for $x$x: |
$x$x |
$=$= | $-2$−2 |

To find the $y$`y` coordinate of the point at $x=-2$`x`=−2 we substitute the $x$`x` value back into the equation for $y$`y`:

$y$y |
$=$= | $3-4x^2$3−4x2 |

$y$y |
$=$= | $3-4(-2)^2$3−4(−2)2 |

$y$y |
$=$= | $-13$−13 |

Therefore, the point on the curve $y=3-4x^2$`y`=3−4`x`2 where the gradient is $16$16 is $(-2,\ -13).$(−2, −13).

Find the equation of the tangent and normal to the curve with equation $y=x^3-3x^2+2$`y`=`x`3−3`x`2+2, at the point $(1,0)$(1,0)

Think: To find the equations of the tangent and the normal we need to know the gradient of the two lines and a single point on each line. The gradient will be given by the derivative of the curve evaluated at that point.

**Do: **Find the gradient function by differentiating $y=x^3-3x^2+2$`y`=`x`3−3`x`2+2:

$y'=3x^2-6x$`y`′=3`x`2−6`x`

Evaluate the gradient at the point $(1,0)$(1,0):

$y'(1)=3(1)^2-6\times1=-3$`y`′(1)=3(1)2−6×1=−3

Identify the gradient of the tangent and the gradient of the normal.

The value of the gradient at the point is $m_1=-3$`m`1=−3 as found above. The gradient of the normal will be:

$m_2=-\frac{1}{m_1}=-\frac{1}{-3}=\frac{1}{3}$`m`2=−1`m`1=−1−3=13

Find the equation of the tangent using the point gradient formula with point $(1,0)$(1,0) and gradient $-3$−3.

$y-y_1$y−y1 |
$=$= | $m(x-x_1)$m(x−x1) |

$y-0$y−0 |
$=$= | $-3(x-1)$−3(x−1) |

$y$y |
$=$= | $-3x+3$−3x+3 |

This is the equation of the tangent.

Find the equation of the normal using the point gradient formula with point $(1,0)$(1,0) and gradient $\frac{1}{3}$13:

$y-y_1$y−y1 |
$=$= | $m(x-x_1)$m(x−x1) |

$y-0$y−0 |
$=$= | $\frac{1}{3}(x-1)$13(x−1) |

$y$y |
$=$= | $\frac{x}{3}-\frac{1}{3}$x3−13 |

This is the equation of the normal.

Let's just confirm that these all look correct on a graph.

Another way to think about the gradient of a line is to look at the angle of inclination, which is the angle that a line makes with the positive $x-axis$`x`−`a``x``i``s`. The following diagram illustrates the angle of inclination$(\theta)$(`θ`) in a right-angled triangle $ABC.$`A``B``C`.

Using right-angled trigonometry and equating to our existing definition of gradient, with the rise $=\ AB$= `A``B` and the run $=\ BC$= `B``C` we can say that:

$\tan(\theta)=\frac{opposite}{adjacent}=\frac{AB}{BC}=\frac{rise}{run}=m$`t``a``n`(`θ`)=`o``p``p``o``s``i``t``e``a``d``j``a``c``e``n``t`=`A``B``B``C`=`r``i``s``e``r``u``n`=`m`

This leads us to the result that:

Formula for finding the gradient using the angle of inclination

$m=\tan(\theta)$`m`=`t``a``n`(`θ`)

To find an angle using $m=\tan(\theta)$`m`=`t``a``n`(`θ`), rearrange the formula such that it becomes

That is:

$\theta=\tan^{-1}(x)$`θ`=`t``a``n`−1(`x`)

Find the gradient of a line with an angle of inclination of $45°$45° with the positive $x-axis$`x`−`a``x``i``s`

**Think:** The formula $m=\tan(\theta)$`m`=`t``a``n`(`θ`) relates the gradient and angle of inclination.

$m=\tan45°$`m`=`t``a``n`45°

$m=1$`m`=1

Therefore, the gradient of the line is $1$1.

By considering the graph of $f\left(x\right)=2x$`f`(`x`)=2`x`, find $f'$`f`′$\left(-5\right)$(−5).

Consider the curve $f\left(x\right)=x^2+8x+15$`f`(`x`)=`x`2+8`x`+15.

Find $f'\left(x\right)$

`f`′(`x`).Find the gradient of the tangent to the curve at the point $\left(4,63\right)$(4,63).

Determine the equation of the tangent to the curve $f\left(x\right)=x^2+8x+15$

`f`(`x`)=`x`2+8`x`+15 at $\left(4,63\right)$(4,63).Express the equation of the tangent line in the form $y=mx+c$

`y`=`m``x`+`c`.Find the gradient of the normal to the curve at the point $\left(4,63\right)$(4,63).

Determine the equation of the normal to the curve $f\left(x\right)=x^2+8x+15$

`f`(`x`)=`x`2+8`x`+15 at $\left(4,63\right)$(4,63).

A line passing through the points $\left(2,-3\right)$(2,−3) and $\left(x,9\right)$(`x`,9) makes an angle of $120^\circ$120° with the positive $x$`x`-axis. Solve for the value of $x$`x`, expressing your answer in simplest rationalised form.

Apply differentiation to gradients, tangents and normals.