# derivative

#### rcs

##### New member
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
If $y = e^{f(x)}$ then $y' = e^{f(x)} f'(x)$. Try that.

#### Prove It

##### Well-known member
MHB Math Helper
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 $\displaystyle \frac{dy}{du}$?

rcs

#### rcs

##### New member
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
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
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.

#### Prove It

##### Well-known member
MHB Math Helper
What's the derivative of $\displaystyle e^x$?

#### rcs

##### New member
What's the derivative of $\displaystyle e^x$?

isn't not e^x also?

#### Jameson

Staff member
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)?

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

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#### rcs

##### New member
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
can anybody show or help me on a step by step on this please?

thanks a lot

#### earboth

##### Active member
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:

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

$g(x) = -{\frac12}x$ 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
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:

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

$g(x) = -{\frac12}x$ 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
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.