can someone please help me with this problem....

.e^(x/2)

----------

..x^(5)

can someone please give me an idea of how to do this problem... thank you!! :confused: :(

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- Jan 21st 2006, 01:04 PMdemi[SOLVED] HELP !! defferentiation exponents
can someone please help me with this problem....

.e^(x/2)

----------

..x^(5)

can someone please give me an idea of how to do this problem... thank you!! :confused: :( - Jan 21st 2006, 01:38 PMCaptainBlackQuote:

Originally Posted by**demi**

$\displaystyle \frac{e^{x/2}}{x^5}$.Well this can be done in a number of ways, you could use the quotient rule,

or the product rule to find this derivative. Now I don't have that good a

memory, and so never bother with the quotient rule, always using the product rule:

$\displaystyle \frac{d}{dx} f(x)g(x)=\frac{df}{dx}g(x)+f(x)\frac{dg}{dx}$.

Here we let:

$\displaystyle f(x)=e^{x/2}$,

and:

$\displaystyle g(x)=\frac{1}{x^5}$.

Now:

$\displaystyle \frac{d}{dx}f=\frac{d}{dx}e^{x/2}=\frac{1}{2}e^{x/2}$,

and:

$\displaystyle \frac{d}{dx}g=\frac{d}{dx}\left\{ \frac{1}{x^5} \right\}=-5\frac{1}{x^6}$.

So now you have enough to assemble the derivative :)

RonL - Jan 21st 2006, 02:12 PMticbolQuote:

Originally Posted by**demi**

d/dx [u/v] = [v*du/dx -u*dv/dx] / v^2 ------------***

d/dx [e^(x/2) / x^5]

= {[(x^5)*e^(x/2) *(1/2)] -[e^(x/2) *(5x^4)]} / [(x^5)^2]

= {(1/2)(x^5)e^(x/2) -5(x^4)e^(x/2)} / x^10

= [e^(x/2)][(x^5 /2) -(5x^4)]/ x^10

= [e^(x/2)][1 /(2x^5) -5/(x^6)] ---------------------(i)

= [e^(x/2)][{[1 /(2x^5)]*(x/x)} -{[5/(x^6)]*(2/2)]}

= [e^(x/2)][x /(2x^6) -10/(2x^6)]

= [e^(x/2)][(x -10) /(2x^6)]

= (1/2)[e^(x/2)][(x -10) /(x^6)] ------------answer.

--------------------------------------

d/dx [u*v] = v*du/dx +u*dv/dx ----------***

d/dx [e^(x/2) / x^5]

= d/dx [(x^(-5))(e^(x/2))]

= e^(x/2) *[-5 x^(-6)] +[x^(-5) *[e^(x/2) *(1/2)]

= [e^(x/2)][-5/(x^6) +(1 /(2x^5)]

= [e^(x/2)][1/(2x^5) -5/(x^6)] ------------same as (i) above.

Etc...