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# Differentiation watch

1. Heya, could somebody check this question
Find the coordinates of the maximum and minimum point on the curve y=x^2(5-x)^3, I got (5,0) and (2, 108) the 108 makes me think im completely wrong.
Also, this question says show that d/dx = something, but what does d/dx mean?
and if I had to sketch a graph of y=(x-1)^2, would it only cross the x axis at 1 and nowhere else?
sorry for all the questions, haha
2. (Original post by jacksonmeg)
Heya, could somebody check this question
Find the coordinates of the maximum and minimum point on the curve y=x^2(5-x)^3, I got (5,0) and (2, 108) the 108 makes me think im completely wrong.
Also, this question says show that d/dx = something, but what does d/dx mean?
and if I had to sketch a graph of y=(x-1)^2, would it only cross the x axis at 1 and nowhere else?
sorry for all the questions, haha
d/dx means differentiate with respect to x what follows.

d/dx of 2x is 2.

Is that x to the power of 2(5-x), or x^2 multiplied by (5-x)?
3. (Original post by Liamnut)
d/dx means differentiate with respect to x what follows.

d/dx of 2x is 2.

Is that x to the power of 2(5-x), or x^2 multiplied by (5-x)?
x^2 multiplied by (5-x)^3
4. (Original post by jacksonmeg)
Heya, could somebody check this question
Find the coordinates of the maximum and minimum point on the curve y=x^2(5-x)^3, I got (5,0) and (2, 108) the 108 makes me think im completely wrong.
Also, this question says show that d/dx = something, but what does d/dx mean?
and if I had to sketch a graph of y=(x-1)^2, would it only cross the x axis at 1 and nowhere else?
sorry for all the questions, haha
If you mean then there are actually 3 stationary points, one of which is a point of inflection.

Stationary points occur where there is 0 gradient i.e.

Could you show us your working?
5. (Original post by Khallil)
If you mean then there are actually 3 stationary points, one of which is a point of inflection.

Stationary points occur where there is 0 gradient i.e.

Could you show us your working?
u= x^2 v = (5-x)^3

du/dx = 2x dv/dx = -3(5-x)^2

-3x^2(5-x)^2 + 2x(5-x)^3
x(5-x)^2(-5x+10) = 0
5-x = 0, x = 5, y = 0
-5x+10 = 0, x = 2, y = 108
6. (Original post by jacksonmeg)
u= x^2 v = (5-x)^3

du/dx = 2x dv/dx = -3(5-x)^2

-3x^2(5-x)^2 + 2x(5-x)^3
x(5-x)^2(-5x+10) = 0
5-x = 0, x = 5, y = 0
-5x+10 = 0, x = 2, y = 108
You seem to be forgetting

To verify that each of these points are maxima, minima or points of inflection, you need to evaluate the second derivative of the function with respect to .
7. (Original post by Khallil)
You seem to be forgetting

To verify that each of these points are maxima, minima or points of inflection, you need to evaluate the second derivative of the function with respect to .
oh yeah.... -idiot- thanks, the rest is ok though?
8. (Original post by jacksonmeg)
oh yeah.... -idiot- thanks, the rest is ok though?
yes
9. (Original post by jacksonmeg)
oh yeah.... -idiot- thanks, the rest is ok though?
Yep!
10. (Original post by Khallil)
Yep!
for the second differentiation, would u = x(5-x)^2 and v = -5x+10 ?
11. (Original post by Khallil)
Yep!
im getting 0, 0 and -90, are the 2 0s the turning points?
12. (Original post by jacksonmeg)
for the second differentiation, would u = x(5-x)^2 and v = -5x+10 ?
Think it's a bit easier to do u= (5-x)^2 v = -5x^2+10x. Then you don't end up using the product rule twice.
13. (Original post by jacksonmeg)
for the second differentiation, would u = x(5-x)^2 and v = -5x+10 ?
Yep! You could even do the product rule with three terms, where , and and functions of :

(Original post by jacksonmeg)
im getting 0 and -90 so i cant of done it right
If those are values of the second derivative, I got the same values. However, you're missing out the second derivative evaluated at
14. (Original post by Khallil)
Yep! You could even do the product rule with three terms, where , and and functions of :

If those are values of the second derivative, I got the same values. However, you're missing out the second derivative evaluated at
yeah I just put 0 in and also got 0, so how would you differentiate between max and min? plot a graph?
15. (Original post by jacksonmeg)
yeah I just put 0 in and also got 0, so how would you differentiate between max and min? plot a graph?
What did you get for
16. (Original post by Khallil)
What did you get for
-5x(5-x)^2 -2x(5-x)(-5x+10)
17. (Original post by jacksonmeg)
-5x(5-x)^2 -2x(5-x)(-5x+10)
Try this as it may be simpler. Your first derivative is correct, so you probably made an error when trying to find the second derivative.

Now find by letting and letting
18. (Original post by Khallil)
Try this as it may be simpler. Your first derivative is correct, so you probably made an error when trying to find the second derivative.

Now find by letting and letting
5(5-x)^2(2-2x) - 10(5-x)(2x-x^2) ?
19. (Original post by jacksonmeg)
5(5-x)^2(2-2x) - 10(5-x)(2x-x^2) ?
Yep!

To make everything neater, it usually helps to factorise out any terms that are common:

Have you heard of the second derivative test to determine the nature of stationary points?
20. (Original post by Khallil)
Yep!

To make everything neater, it usually helps to factorise out any terms that are common:

Have you heard of the second derivative test to determine the nature of stationary points?
dont you put in the x values and the positive one is minimum, negative is max? I figured it didnt matter about simplifying since you just need to put in x values, and i usually go wrong anyway. lol

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