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derivative

rcs

New member
Feb 5, 2012
10
I am so happy that i found this site because it's been few days now that i haven't solved this function.
I hope somebody here can help me on finding the derivative of this function..


Thank you so much.
 

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Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
If $y = e^{f(x)}$ then $y' = e^{f(x)} f'(x) $. Try that.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
I am so happy that i found this site because it's been few days now that i haven't solved this function.
I hope somebody here can help me on finding the derivative of this function..


Thank you so much.
What Danny is using is the Chain Rule. The more general result of the chain rule is that if $ \displaystyle y = y \left( u(x) \right) $ (a function of a function), then $ \displaystyle \frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dy}{du} $.

So in your case, you have $ \displaystyle y = 2e^{-\frac{x}{2}\cos{\left( -\frac{x}{2} \right)}} $, so let $ \displaystyle u = -\frac{x}{2}\cos{\left( - \frac{x}{2} \right)} $ so that $ \displaystyle y = 2e^u $. Can you now evaluate $ \displaystyle \frac{du}{dx} $ and $ \displaystyle \frac{dy}{du} $?
 
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rcs

New member
Feb 5, 2012
10
sir,

do u mean that i have to get the derivative first of the whatever is y = 2e^u

does that ln y = ln e^2u?

then ln y = 2u remain and manipulate for the derivate of 2u?

hope u i am doing it correctly


Thank you
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
sir,

do u mean that i have to get the derivative first of the whatever is y = 2e^u

does that ln y = ln e^2u?

then ln y = 2u remain and manipulate for the derivate of 2u?

hope u i am doing it correctly


Thank you
No, that's incorrect. Say we had $ \displaystyle y = 2e^x $, what would $ \displaystyle \frac{dy}{dx} $ be?

To find $ \displaystyle \frac{dy}{du} $ from $ \displaystyle y = 2e^u $ is identical, it's just that the function is a function of u instead of x, and the derivative is taken with respect to u instead of x.
 

rcs

New member
Feb 5, 2012
10
No, that's incorrect. Say we had $ \displaystyle y = 2e^x $, what would $ \displaystyle \frac{dy}{dx} $ be?

To find $ \displaystyle \frac{dy}{du} $ from $ \displaystyle y = 2e^u $ is identical, it's just that the function is a function of u instead of x, and the derivative is taken with respect to u instead of x.

:( made me so confuse
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403

rcs

New member
Feb 5, 2012
10

Jameson

Administrator
Staff member
Jan 26, 2012
4,035
Correct! Now consider the case of $ \displaystyle e^{f(x)}$. The derivative is $ \displaystyle e^{f(x)}*f'(x)$. What is f(x) in this problem and what is the derivative of f(x)?

$ \displaystyle f(x)= -\frac{x}{2}\cos{\left( - \frac{x}{2} \right)} $
 
Last edited:

rcs

New member
Feb 5, 2012
10
Correct! Now consider the case of $ \displaystyle e^{f(x)}$. The derivative is $ \displaystyle e^{f(x)}*f'(x)$. What is f(x) in this problem and what is the derivative of f(x)?

$ \displaystyle f(x)= -\frac{x}{2}\cos{\left( - \frac{x}{2} \right)} $

that is my problem now... i have a hard of solving trigonometry :( so hard for me on that ...
 

rcs

New member
Feb 5, 2012
10
can anybody show or help me on a step by step on this please?

thanks a lot
 

earboth

Active member
Jan 30, 2012
74
can anybody show or help me on a step by step on this please?

thanks a lot
One personal remark first: I don't like the idea that we should do your work ...

1. You must see at once that
$ f(x)=2 e^{-{\tfrac12}x \cdot \cos\left(-{\tfrac12}x \right)} $ is an exponential function.

What function do you get if you differentiate $ f(x)=e^x $ ?

2. You must see at once that your function is not a simple exponential function. The exponent is a function itself. Thus you have to use the chain rule when differentiating your function:
$ f(x) = e^{u(x)}~\implies~ f'(x) = e^{u(x)} \cdot u'(x) $

3. You must see at once that the exponent of your function is a product of functions. To differentiate a product of functions you have to use the product rule:

$ f(x) = g(x) \cdot h(x) ~ \implies~ f'(x) = h(x) \cdot g'(x) + g(x) \cdot h'(x) $

With your function you have:
$ g(x) = -{\frac12}x $ and
$ h(x) = \cos\left( -\frac12 x \right) $

4. You are supposed to know that
$ \sin'(x) = \cos(x) $ ............$ \cos'(x) = -\sin(x) $ ............ $ -\sin'(x) = -\cos(x) $ ............$ -\cos'(x) = \sin(x) $ ............
That means during the differentiation of the exponent you have to differentiate the cos-function. BUT: the argument of the cos-function isn't a simple x therefore you must use the chain rule here:

$ f(x) = \cos(u(x))~\implies~f'(x) = -\sin(u(x)) \cdot u'(x) $

5. Now collect all the results to one brand-new function.
 

rcs

New member
Feb 5, 2012
10
One personal remark first: I don't like the idea that we should do your work ...

1. You must see at once that
$ f(x)=2 e^{-{\tfrac12}x \cdot \cos\left(-{\tfrac12}x \right)} $ is an exponential function.

What function do you get if you differentiate $ f(x)=e^x $ ?

2. You must see at once that your function is not a simple exponential function. The exponent is a function itself. Thus you have to use the chain rule when differentiating your function:
$ f(x) = e^{u(x)}~\implies~ f'(x) = e^{u(x)} \cdot u'(x) $

3. You must see at once that the exponent of your function is a product of functions. To differentiate a product of functions you have to use the product rule:

$ f(x) = g(x) \cdot h(x) ~ \implies~ f'(x) = h(x) \cdot g'(x) + g(x) \cdot h'(x) $

With your function you have:
$ g(x) = -{\frac12}x $ and
$ h(x) = \cos\left( -\frac12 x \right) $

4. You are supposed to know that
$ \sin'(x) = \cos(x) $ ............$ \cos'(x) = -\sin(x) $ ............ $ -\sin'(x) = -\cos(x) $ ............$ -\cos'(x) = \sin(x) $ ............
That means during the differentiation of the exponent you have to differentiate the cos-function. BUT: the argument of the cos-function isn't a simple x therefore you must use the chain rule here:

$ f(x) = \cos(u(x))~\implies~f'(x) = -\sin(u(x)) \cdot u'(x) $

5. Now collect all the results to one brand-new function.

Thank you so much... i only thought of the use of term help in this website... this is the significance of this website now... because there are some instances that some people really don't know that problem and seek for help...

thanks so much!
 

sweer6

New member
May 21, 2014
10
if y = e^(f(x) then dy/dx = f'(x)*e^(f(x))

this means to multiply the original function (e^x) by the deravative of x which is positive 1 (+1), to give itself as the differential. The differential of e^x = e^x. (this is pretty cool)
or y = e^x then dy/dx = e^x.