Answer:
Differentiation can be used to find the gradient of the graph at a particular point.
To differentiate,
- bring down the power
- power minus 1
Differentiating y with respect to x term would be written as [tex] \frac{dy}{dx} [/tex]
Bringing down the power means to multiply the digit in the power to the equation first.
Let me show you an example.
[tex]y = 5x^{2} [/tex]
[tex] \frac{dy}{dx} = 2(5x^{2 - 1} ) \\ \frac{dy}{dx} = 10x^{1} \\ \frac{dy}{dx} = 10x[/tex]
Hence the gradient of the graph y=5x² at any point is given by the formula of 10x.
So at x=2,
gradient of graph is 10(2)= 20
Knowing the gradient, you can now find 2 points to plot your tangent using the gradient formula:
[tex] \frac{y1 - y2}{x1 - x2} [/tex]
Let's look at example 3 again.
In this case, the graph is y=x³.
① Differentiate y with respect to x
[tex] \frac{dy}{dx} = 3(x^{3 - 1} ) \\ \frac{dy}{dx} = 3 {x}^{2} [/tex]
② Find the value of the gradient at the point given.
When x=2,
[tex] \frac{dy}{dx} = 3(2)^{2} \\ \frac{dy}{dx} = 3(4) \\ \frac{dy}{dx} = 12[/tex]
This means that the gradient of the tangent at x=2 is 12.
So at x=2, y=8
Let y2 be 8 and x2 be 2.
Subst into the gradient formula:
[tex] \frac{y1 - 8}{x1 - 2} = 12 \\ y1 - 8 = 12(x1 - 2) \\ y1 - 8 = 12x1 - 24 \\ y1 = 12x1 - 24 + 8 \\ y1 = 12x1 - 16[/tex]
Now you have an equation which shows how any coordinates that lie on the tangent are related to each other.
Use this to find 2 coordinates so you can draw a line through them to draw the tangent.
When x=3,
y= 12(3) -16
y= 36 -16
y= 20
when x= 1.5,
y= 12(1.5) -16
y= 2
Plot these 2 points: (3 ,20) and (1.5, 2)
Now draw the tangent through these 2 points.
Let's check:
[tex]gradient = \frac{20 - 2}{3 - 1.5} \\ gradeint = \frac{18}{1.5} \\ gradient = 12[/tex]
Essentially, what we are trying to achieve is an accurate gradient value. However, we are not allowed to use differentiation ( Sec 4 A Math syllabus ) to answer the question. So use that as a side working to achieve an accurate tangent or you can also use that to check if your answer is close to the correct answer.
If the equation is y= x³-12 instead,
note that you have to differentiate x³ just like in example 3, ignoring the -12 since differentiating a constant would give you zero.
Further explanation:
Constants are basically _x⁰.
3= 3x⁰
4= 4x⁰
This is because x⁰= 1.
So if we differentiate 3,
we get d/dx (3x⁰)= 0(3x^-1)
This would give us 0 since anything multiplied by 0 is still 0.
