Differentiate e^(-1/2 x) sqrt(1 + 2x) - troubles!

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  • #1
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the curve y = e^ −1/2x √(1 + 2x) please differentiate this equation i am having some problem.. i want to check my answer

My answer :

e^-1/2x { (1+2x )^-1/2 + (1+2x)^1/2 (-1/2) }

In fact i need to find the x-coordinates and according the booklet the answer of the x-coordinate is 0.5 and me i cannot have the answer ( need to equal the dy/dx = 0 ) to find x-coordinate
 

Answers and Replies

  • #2
ehild
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Do you need to find the value of x where dy/dx=0? You derived the equation already, put it equal to 0 and solve. As the exponent can not be zero, the factor { (1+2x )^-1/2 + (1+2x)^1/2 (-1/2) }=0.

ehild
 
  • #3
Mentallic
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differentiation of two products means you must use the product rule.

if y=uv, where u and v are functions of x, such as in your case you have [itex]u=e^{-x/2}[/itex] and [itex]v=(1+2x)^{1/2}[/itex]

then [tex]\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}[/tex]

Ok so what is [tex]\frac{dv}{dx}[/tex] and what is [tex]\frac{du}{dx}[/tex] ?

Solve these separately first and we'll show you if you made any mistakes anywhere in your differentiation.
 
  • #4
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differentiation of two products means you must use the product rule.

if y=uv, where u and v are functions of x, such as in your case you have [itex]u=e^{-x/2}[/itex] and [itex]v=(1+2x)^{1/2}[/itex]

then [tex]\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}[/tex]

Ok so what is [tex]\frac{dv}{dx}[/tex] and what is [tex]\frac{du}{dx}[/tex] ?

Solve these separately first and we'll show you if you made any mistakes anywhere in your differentiation.

Okk so for U = e^-1/2x...........du/dx = -1/2 e^-1/2x

and for V = (1+2x)^1/2..........dv/dx = (1+2x)^-1/2

and i replace in y=uv
 
  • #5
Mentallic
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Nearly. For [tex]v=(1+2x)^{1/2}[/tex] you need to remember to take the derivative of the inside of the brackets by the chain rule. For [tex]y=[f(x)]^n[/tex], [tex]\frac{dy}{dx}=n[f(x)]^{n-1}.f'(x)[/tex]

And then you replace in [tex]\frac{dy}{dx}=uv'+vu'[/tex] where v' is the derivative of v with respect to x, or [tex]\frac{dv}{dx}[/tex] and similarly for u'.
 
  • #6
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Nearly. For [tex]v=(1+2x)^{1/2}[/tex] you need to remember to take the derivative of the inside of the brackets by the chain rule. For [tex]y=[f(x)]^n[/tex], [tex]\frac{dy}{dx}=n[f(x)]^{n-1}.f'(x)[/tex]

And then you replace in [tex]\frac{dy}{dx}=uv'+vu'[/tex] where v' is the derivative of v with respect to x, or [tex]\frac{dv}{dx}[/tex] and similarly for u'.


no i have already evalute,, 1/2 (1+2x) ^-1/2 x 2 = (1 +2x) ^-1/2
 
  • #7
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Do you need to find the value of x where dy/dx=0? You derived the equation already, put it equal to 0 and solve. As the exponent can not be zero, the factor { (1+2x )^-1/2 + (1+2x)^1/2 (-1/2) }=0.

ehild


hmmm no let the exponential and x-coordinate part,, help me to differentiate, infact i have differentiate, but i am not obtaining the answer
 
  • #8
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yeahh infact the x-coordinate of a maximum point
 
  • #9
ehild
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You have differentiate the y(x) function already. Now you have to solve the equation

(1+2x )^-1/2 + (1+2x)^1/2 (-1/2)=0

I rewrite it in tex so as you see more clearly:

[tex]\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0[/tex]

hint: Multiply the equation with (1+2x) ^1/2.

ehild
 
  • #10
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You have differentiate the y(x) function already. Now you have to solve the equation

(1+2x )^-1/2 + (1+2x)^1/2 (-1/2)=0

I rewrite it in tex so as you see more clearly:

[tex]\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0[/tex]

hint: Multiply the equation with (1+2x) ^1/2.

ehild

[tex]\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0[/tex]

Till here i have reached, so

1 / (1+2x)^1/2 = 1/2 (1+2x)^1/2

If i use your hint i get: 0 = 1/2 (1+2X) ^1/2 x (1+2X) ^1/2

Aww now how should i break the brackets ?
 
  • #11
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## Oppps i think that it is not zero but one
 
  • #12
ehild
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[tex]\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0[/tex]

Till here i have reached, so

1 / (1+2x)^1/2 = 1/2 (1+2x)^1/2

If i use your hint i get: 0 = 1/2 (1+2X) ^1/2 x (1+2X) ^1/2

Aww now how should i break the brackets ?

Your last equation is totally wrong. I said multiply the whole equation by
(1+2X) ^1/2. Do it with your first equation, it is correct.

ehild
 
  • #13
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Your last equation is totally wrong. I said multiply the whole equation by
(1+2X) ^1/2. Do it with your first equation, it is correct.

ehild


so the equation is like this after the multiplication:

1= 1/2 (1+2X)^1/2 X (1+2X) ^1/2

1 = 1/2 (1 + 2x)

is it correct?
 
  • #14
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by the way the answer that is the x-coordinate is 0.5 according to the booklet
 
  • #15
HallsofIvy
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so the equation is like this after the multiplication:

1= 1/2 (1+2X)^1/2 X (1+2X) ^1/2

1 = 1/2 (1 + 2x)

is it correct?
Yes, that is correct. And that is a simple linear equation. Solve it for x.
 

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