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3) JUNE 2013 4H Q. 11

(c) For the curve with equation [tex]$y = x^3 - 6x^2 + 20$[/tex]:

(i) Find [tex]\frac{dy}{dx}[/tex].

(ii) Find the gradient of the curve at [tex]x = -3[/tex].


Sagot :

Let's address the problem step by step.

The curve is defined by the equation [tex]\( y = x^3 - 6x^2 + 20 \)[/tex].

### (i) Finding the derivative [tex]\(\frac{d y}{d x}\)[/tex]

To find the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex], we use the rules of differentiation. The curve equation given is:

[tex]\[ y = x^3 - 6x^2 + 20 \][/tex]

Using the power rule of differentiation, which states that [tex]\(\frac{d}{dx}[x^n] = nx^{n-1}\)[/tex], we differentiate each term with respect to [tex]\( x \)[/tex]:

1. The derivative of [tex]\( x^3 \)[/tex] is [tex]\( 3x^2 \)[/tex].
2. The derivative of [tex]\( -6x^2 \)[/tex] is [tex]\(-6 \times 2x = -12x \)[/tex].
3. The derivative of the constant [tex]\( 20 \)[/tex] is [tex]\( 0 \)[/tex].

Combining these, we get the derivative:

[tex]\[ \frac{d y}{d x} = 3x^2 - 12x \][/tex]

### (ii) Finding the gradient of the curve at [tex]\( x = -3 \)[/tex]

The gradient of the curve at a specific point [tex]\( x = -3 \)[/tex] is given by the value of the derivative [tex]\(\frac{d y}{d x}\)[/tex] at that point.

We substitute [tex]\( x = -3 \)[/tex] into the derivative we found in part (i):

[tex]\[ \frac{d y}{d x} \Bigg|_{x = -3} = 3(-3)^2 - 12(-3) \][/tex]

Now calculate it step-by-step:

1. [tex]\((-3)^2 = 9 \)[/tex]
2. [tex]\( 3 \times 9 = 27 \)[/tex]
3. [tex]\( 12 \times -3 = -36 \)[/tex]
4. Therefore, [tex]\( 27 + 36 = 63 \)[/tex]

So the gradient of the curve at [tex]\( x = -3 \)[/tex] is [tex]\( 63 \)[/tex].

### Final solutions:

- The derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 3x^2 - 12x \)[/tex].
- The gradient of the curve at [tex]\( x = -3 \)[/tex] is [tex]\( 63 \)[/tex].
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