Mathematics, 08.04.2020 09:11 alyvia05

Hi:) my teacher taught another method but she said we’ll only be learning it next year. I didn’t understand it , anyone able to explain? thank you so much! :) ♡

Hi:) my teacher taught another method but she said we’ll only be learning it next year. I didn’t und

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Answer from: rakanmadi87

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 $\frac{dy}{dx}$

Bringing down the power means to multiply the digit in the power to the equation first.

Let me show you an example.

$y = 5x^{2}$

$\frac{dy}{dx} = 2(5x^{2 - 1} ) \\ \frac{dy}{dx} = 10x^{1} \\ \frac{dy}{dx} = 10x$

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:

$\frac{y1 - y2}{x1 - x2}$

Let's look at example 3 again.

In this case, the graph is y=x³.

① Differentiate y with respect to x

$\frac{dy}{dx} = 3(x^{3 - 1} ) \\ \frac{dy}{dx} = 3 {x}^{2}$

② Find the value of the gradient at the point given.

When x=2,

$\frac{dy}{dx} = 3(2)^{2} \\ \frac{dy}{dx} = 3(4) \\ \frac{dy}{dx} = 12$

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:

$\frac{y1 - 8}{x1 - 2} = 12 \\ y1 - 8 = 12(x1 - 2) \\ y1 - 8 = 12x1 - 24 \\ y1 = 12x1 - 24 + 8 \\ y1 = 12x1 - 16$

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:

$gradient = \frac{20 - 2}{3 - 1.5} \\ gradeint = \frac{18}{1.5} \\ gradient = 12$

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.

Answer from: Quest

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Answer from: Quest

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