Edexcel c3 differentiation - help

Original post by iAre Teh Lejend

I've done part a already, and got dy/dx = e^-x (x^3 + 3x^2 - 2x - 2) and now im stuck on b.

I've not looked at the rest of it, but your dy/dx is incorrect. There should be a minus in front of the x^3 and it should be +2x, not -2x, or are they just typos?

Reply 2

OP

Ohh, yes you're right. Sorry, I was rushing the part where I took the common factor e^-x out, let me try again and I'll post if I get the correct answer :biggrin:

Reply 3

OP

You were right for part a, but that in the end didn't affect part b as the gradient for the tangent remained as -2. Still stuck on part b :frown:

Reply 4

Original post by iAre Teh Lejend

Now to find x-coordinate at point P, I did the following;

x^3 - 2x = 1/2x

What happened to the e^(-x) on the LHS? This is presumably meant to be f(x) = x/2.

Reply 5

OP

Original post by ghostwalker

What happened to the e^(-x) on the LHS? This is presumably meant to be f(x) = x/2.

lol I thought you just get rid of it... ( I'm self teaching this part so go easy on me LOOL )

(edited 12 years ago)

Reply 6

Original post by iAre Teh Lejend

lol I thought you just get rid of it... ( I'm self teaching C3 so go easy on me LOOL )

You don't just chuck away terms without good reason, and here there is none. f(x) is defined and that's what you're putting equal to x/2.

Put it back in and rearrange, and you'll get your desired equation.

Reply 7

OP

Okay thanks, let me try... But how do I solve x for;

x/2 = (x^3 - 2x)e^-x ??

(edited 12 years ago)

Reply 8

Original post by iAre Teh Lejend

Okay thanks, let me try... But how do I solve x for;

x/2 = (x^3 - 2x)e^-x ??

Read the question again. You're not asked to solve for x.

Reply 9

OP

Original post by ghostwalker

Read the question again. You're not asked to solve for x.

Aren't I ? But I need to find the x value at point P so I can show that the x value at point P is the solution of the equation 2x^2 = e^x + 4 .... So how do I find the x value then.

Reply 10

OP

When I found dy/dx = e^-x ( -x^3 + 2x + 3x^2 -2 )
I knew that to find the x co at P, I need the eequation of the normal. To find the equation of the normal, I found the gradient of the tangent of C at O, so I subbed x as 0 in dy/dx, to get the gradient as -2. The tangent to the normal at O is perp, so gradient of normal is 1/2. Now we can find equation of normal by using y - y1 = m ( x -x1 ), so equation for normal is y = 1/2x
Now this is where im stuck, to find x co at P, equal equation of C to normal, so we have

1/2x = (x^3 - 2x)e^-x

Reply 11

Original post by iAre Teh Lejend

When I found dy/dx = e^-x ( -x^3 + 2x + 3x^2 -2 )
I knew that to find the x co at P, I need the eequation of the normal. To find the equation of the normal, I found the gradient of the tangent of C at O, so I subbed x as 0 in dy/dx, to get the gradient as -2. The tangent to the normal at O is perp, so gradient of normal is 1/2. Now we can find equation of normal by using y - y1 = m ( x -x1 ), so equation for normal is y = 1/2x
Now this is where im stuck, to find x co at P, equal equation of C to normal, so we have

1/2x = (x^3 - 2x)e^-x

And the x value that is the solution to that equation is the x co-ordinate of the point P. You just need to re-arrange that equation so that it is the same as the required one.

(edited 12 years ago)

Reply 12

Extricated

(x^3-2x)e^-x = 1/2x

note that : e^-x = 1/e^x

multiply both sides by e^x
you get (x^3-2x) = (e^x/2)x

multiply both sides by x

x^4-2x^2 = (e^x/2)x^2
factorise left hand side
x^2(x^2-2) = (e^x/2)x^2
cancel x^2 from both sides

x^2 - 2 = (e^x/2)

2x^2-4 = (E^x)

2x^2 = e^x+4

as required

(edited 12 years ago)

Reply 13

OP