first and second derivatives of these equations

Limits, differentiation, related rates, integration, trig integrals, etc.
doodoo21
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first and second derivatives of these equations

Hi Guys,

I'm having trouble working out some equations mostly dealing with exponential and logs.

the equations are:

y= 2x + 3^(-12x)

y= x^(-2x) - (e^2x)/5

I need to find both the first and second derivative. Any help and explanation would be much appreciated.

Thanks guys.

stapel_eliz
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doodoo21 wrote:y= 2x + 3^(-12x)

y= x^(-2x) - (e^2x)/5

I need to find both the first and second derivative.

Where are you getting stuck?

doodoo21
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Joined: Tue Aug 24, 2010 5:22 am
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Re:

stapel_eliz wrote:
doodoo21 wrote:y= 2x + 3^(-12x)

y= x^(-2x) - (e^2x)/5

I need to find both the first and second derivative.

Where are you getting stuck?

From my understanding the first one has to do with ln

so dy/dx = 2 - 12ln3(3^-12x)

I'm not sure whether I multiply the -12ln3 with the (3^-12x). Even if that is correct or not, I don't understand how to get the second derivative as I'm not sure on how to derive logs >.<

The second one would incorporate the chain rule?

Therefore for x^-2x:

let y= x^u , u = -2x

dy/dx = u(x) . -2 = 4x^2

And for - (e^2x)/5:

let y= - (e^u)/5 , u= 2x

dy/dx = (e^u)/5 . 2 = (2e^2x)/5

I'm not sure on how to handle the /5 and on how to derive the first derivative, especially when they deal with exponential and logs.

If you could further explain where I've gone wrong, I would be grateful.

Martingale
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Re: first and second derivatives of these equations

Let me give you an example...

let $y=x^x$

I want to find $\frac{dy}{dx}$

first I take the log (natural log) of both sides

$\log(y)=\log(x^x)=x\log(x)$

now take the derivative of both sides

$\frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\log(x)=1+\log(x)$

then

$\frac{dy}{dx}=y[1+\log(x)]$

now replace $y$

$\frac{dy}{dx}=x^x[1+\log(x)]$

stapel_eliz
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doodoo21 wrote:I don't understand how to get the second derivative as I'm not sure on how to derive logs

Ouch! They were supposed to teach you that stuff before assigning homework on it!

To learn how to work with exponential and logarithmic functions and their derivatives, try here.

doodoo21 wrote:The second one would incorporate the chain rule?

You can not apply rules for constants to situations with variables. You have the Power Rule for differentiating xn where n is a number; you have the exponential derivative rule for differentiating nx where n is a number. You cannot use either of those rules for xf(x).

Have you learned about logarithmic differentiation yet?

doodoo21
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Re:

stapel_eliz wrote:Ouch! They were supposed to teach you that stuff before assigning homework on it!

To learn how to work with exponential and logarithmic functions and their derivatives, try here.

Thanks for helping. Im doing a master of finance course with no background knowledge on calculus. The lecturer/tutor expected us to know it all so he powered through it :S...Thats why I am in need of assistance!

I would like to know if dy/dx of the first equation is 2 - 12ln3(3^-12x)...I'm still unsure on how it's meant to come out.

stapel_eliz wrote:You can not apply rules for constants to situations with variables. You have the Power Rule for differentiating xn where n is a number; you have the exponential derivative rule for differentiating nx where n is a number. You cannot use either of those rules for xf(x).

Have you learned about logarithmic differentiation yet?

Thank you again...from my understanding dy/dx of the second equation would be: (x^-2x)[ln(x)-2] - (2e^2x)/5

Is this all correct? Thank you again guys.

stapel_eliz
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doodoo21 wrote:I would like to know if dy/dx of the first equation is 2 - 12ln3(3^-12x).

Sorry, but no.

doodoo21 wrote:from my understanding dy/dx of the second equation would be: (x^-2x)[ln(x)-2] - (2e^2x)/5

You're close, but not quite right.