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# C3 Numerical Methods Question watch

1. I'm stuck on this question...
The curve with equation crosses the x-axis at the point .

a. Work out the value of p?
That's

b. The normal to the curve at the point Q, with x-coordinates q, passes through the origin.
Show that is a solution of the equation

This is where I'm stuck. I guess you'd have to work out the gradient of the curve, but how do you differentiate y = ln(3x)? It's not in my book.
That's my first question.
Then after that, you'd find the gradient of the normal, and then work out its equation.
Following that, you'd see where the normal and the curve intersect to work out the value of Q?

Am I right? Thanks!
2. (Original post by Narik)
I'm stuck on this question...
The curve with equation crosses the x-axis at the point .

a. Work out the value of p?
That's

b. The normal to the curve at the point Q, with x-coordinates q, passes through the origin.
Show that is a solution of the equation

This is where I'm stuck. I guess you'd have to work out the gradient of the curve, but how do you differentiate y = ln(3x)? It's not in my book.
That's my first question.
Then after that, you'd find the gradient of the normal, and then work out its equation.
Following that, you'd see where the normal and the curve intersect to work out the value of Q?

Am I right? Thanks!
Y= ln3X, use the chain rule, see if you can do that part.
3. (Original post by JumpingJonny)
Y= ln3X, use the chain rule, see if you can do that part.
In my Edexcel C3 book, this question is in Chapter 4, whereas the chain rule comes later on in the last chapter, which is Chapter 8. Is there any other way that it can be done or is the chain rule the only way? Thanks!
4. For your first question,
the derivative of ln(u) for all y>0 is u'/u

Therefore, the derivative of ln(3x) would be (3x)'/(3x) = 1/x for all x>0

An other way to see it is to write it as ln(3x)=ln(3)+ln(x)
Here, ln3 is just a constant (around 1,098 if you want) therefore its derivative is 0
And then you know that the derivative of ln(x) is 1/x, so by sum, the derivative is 0+1/x=1/x
5. (Original post by Narik)
In my Edexcel C3 book, this question is in Chapter 4, whereas the chain rule comes later on in the last chapter, which is Chapter 8. Is there any other way that it can be done or is the chain rule the only way? Thanks!
The chain rule is the easiest way and would probably take about 10 minutes to learn, isnt the numerical methods coursework designed to be done once you have covered the C3 material already?
6. (Original post by JumpingJonny)
The chain rule is the easiest way and would probably take about 10 minutes to learn, isnt the numerical methods coursework designed to be done once you have covered the C3 material already?
There is no coursework. I'm not an international student if that makes any sense.
And the numerical methods chapter comes before the chain rule. This is an actual exam question, and therefore it assumes you've already covered all the topics. I'll skip it for now then.

Thanks for your help!
7. (Original post by Narik)
I'm stuck on this question...
The curve with equation crosses the x-axis at the point .

a. Work out the value of p?
That's

b. The normal to the curve at the point Q, with x-coordinates q, passes through the origin.
Show that is a solution of the equation

This is where I'm stuck. I guess you'd have to work out the gradient of the curve, but how do you differentiate y = ln(3x)? It's not in my book.
That's my first question.
Then after that, you'd find the gradient of the normal, and then work out its equation.
Following that, you'd see where the normal and the curve intersect to work out the value of Q?

Am I right? Thanks!
you do NOT need the chain rule to calculate the derivative of y = ln (3x).

y = ln (3x) = ln 3 + ln x
therefore dy/dx = d(ln 3)/dx + d(ln x)/dx = 1/x

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