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Differentiation

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

(edited 10 years ago)

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Reply 1

Liamnut

Original post by jacksonmeg

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)?

Reply 2

jacksonmeg

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

Reply 3

user2020user

Original post by jacksonmeg

If you mean

y=x^2 (5-x)^3

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.

\frac{dy}{dx} = 0

Could you show us your working?

Reply 4

jacksonmeg

Original post by Khallil

If you mean

y=x^2 (5-x)^3

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.

\frac{dy}{dx} = 0

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

Reply 5

user2020user

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

x=0

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

x

Reply 6

jacksonmeg

Original post by Khallil

You seem to be forgetting

x=0

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

x

oh yeah.... -idiot- thanks, the rest is ok though?

Reply 7

Liamnut

Original post by jacksonmeg

oh yeah.... -idiot- thanks, the rest is ok though?

yes

Reply 8

user2020user

Original post by jacksonmeg

oh yeah.... -idiot- thanks, the rest is ok though?

Yep!

Reply 9

jacksonmeg

Original post by Khallil

Yep!

for the second differentiation, would u = x(5-x)^2 and v = -5x+10 ?

Reply 10

jacksonmeg

Original post by Khallil

Yep!

im getting 0, 0 and -90, are the 2 0s the turning points?

(edited 10 years ago)

Reply 11

BankOfPigs

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.

Reply 12

user2020user

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

f

g

and

h

and functions of

x

\dfrac{d}{dx} \left( fgh \right) = f'gh + fg'h + fgh'

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

x=0

(edited 10 years ago)

Reply 13

jacksonmeg

Original post by Khallil

Yep! You could even do the product rule with three terms, where

f

g

and

h

and functions of

x

\dfrac{d}{dx} \left( fgh \right) = f'gh + fg'h + fgh'

If those are values of the second derivative, I got the same values. However, you're missing out the second derivative evaluated at

x=0

yeah I just put 0 in and also got 0, so how would you differentiate between max and min? plot a graph?

Reply 14

user2020user

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

\dfrac{d^2y}{dx^2} \ ?

Reply 15

jacksonmeg

Original post by Khallil

What did you get for

\dfrac{d^2y}{dx^2} \ ?

-5x(5-x)^2 -2x(5-x)(-5x+10)

Reply 16

user2020user

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.

\begin{aligned} \dfrac{dy}{dx} & = x(5-x)^2(-5x+10) \\ & = 5x(5-x)^2(2-x) \\ & = 5(5-x)^2(2x-x^2) \end{aligned}

Now find

\dfrac{d^2y}{dx^2}

by letting

u=(5-x)^2

and letting

v=(2x-x^2)

(edited 10 years ago)

Reply 17

jacksonmeg

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.

\begin{aligned} \dfrac{dy}{dx} & = x(5-x)^2(-5x+10) \\ & = 5x(5-x)^2(2-x) \\ & = 5(5-x)^2(2x-x^2) \end{aligned}

Now find

\dfrac{d^2y}{dx^2}

by letting

u=(5-x)^2

and letting

v=(2x-x^2)

5(5-x)^2(2-2x) - 10(5-x)(2x-x^2) ?

Reply 18

user2020user

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:

\dfrac{d^2y}{dx^2} = 10(5-x)(2x^2 - 8x + 5)

Have you heard of the second derivative test to determine the nature of stationary points?

Reply 19

jacksonmeg

Original post by Khallil

Yep!

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

\dfrac{d^2y}{dx^2} = 10(5-x)(2x^2 - 8x + 5)

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