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

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

ehild
Homework Helper

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

Mentallic
Homework Helper

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 $u=e^{-x/2}$ and $v=(1+2x)^{1/2}$

then $$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$$

Ok so what is $$\frac{dv}{dx}$$ and what is $$\frac{du}{dx}$$ ?

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

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 $u=e^{-x/2}$ and $v=(1+2x)^{1/2}$

then $$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$$

Ok so what is $$\frac{dv}{dx}$$ and what is $$\frac{du}{dx}$$ ?

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

Mentallic
Homework Helper

Nearly. For $$v=(1+2x)^{1/2}$$ you need to remember to take the derivative of the inside of the brackets by the chain rule. For $$y=[f(x)]^n$$, $$\frac{dy}{dx}=n[f(x)]^{n-1}.f'(x)$$

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

Nearly. For $$v=(1+2x)^{1/2}$$ you need to remember to take the derivative of the inside of the brackets by the chain rule. For $$y=[f(x)]^n$$, $$\frac{dy}{dx}=n[f(x)]^{n-1}.f'(x)$$

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

no i have already evalute,, 1/2 (1+2x) ^-1/2 x 2 = (1 +2x) ^-1/2

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

yeahh infact the x-coordinate of a maximum point

ehild
Homework Helper

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:

$$\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0$$

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

ehild

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:

$$\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0$$

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

ehild

$$\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0$$

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 ?

## Oppps i think that it is not zero but one

ehild
Homework Helper

$$\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0$$

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

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?

by the way the answer that is the x-coordinate is 0.5 according to the booklet

HallsofIvy
Homework Helper

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.