differentiation applying two rules

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June 28th 2011, 05:50 AM
vidhi96

differentiation applying two rules

Hi I am fine when it comes to straightforward differntiation but when its composite I mean when you have to apply say product rule and then chain rule etc. I really haven't got a clue when to proceed with second rule. I have a problem here and I will really appriciate it if someone can explain in details:

f(x)=e^(6x) sq.rt.(1+3x^2) I don't know how to use fancy text but this should read
f(x) = e to the power of 6x multiplied by square root of 1 plus 3x squared.

I know you apply prodcut and then composite rule- but I get stuck as soon as I did prodcut rule. if someone can illustarte the whole process and logic that will make my life easier
June 28th 2011, 05:55 AM
Archie Meade

Re: differentiation applying two rules

Quote:

Originally Posted by vidhi96

Hi I am fine when it comes to straightforward differntiation but when its composite I mean when you have to apply say product rule and then chain rule etc. I really haven't got a clue when to proceed with second rule. I have a problem here and I will really appriciate it if someone can explain in details:

f(x)=e^(6x) sq.rt.(1+3x^2) I don't know how to use fancy text but this should read
f(x) = e to the power of 6x multiplied by square root of 1 plus 3x squared.

I know you apply prodcut and then composite rule- but I get stuck as soon as I did prodcut rule. if someone can illustarte the whole process and logic that will make my life easier

The Product Rule is

$(uv)'=u'(v)+v'(u)$

Hence you write your derivatives for u and v
and plug them into the formula.

This will be relatively easy if you can get those derivatives,
which require the Chain Rule.

How is this so far ?
June 28th 2011, 05:58 AM
Prove It

Re: differentiation applying two rules

$\displaystyle \begin{align*}y &= e^{6x}(1 + 3x^2)^{\frac{1}{2}} \\ \frac{dy}{dx} &= e^{6x}\frac{d}{dx}\left[(1 + 3x^2)^{\frac{1}{2}}\right] + (1 + 3x^2)^{\frac{1}{2}}\frac{d}{dx}\left(e^{6x}\right) \end{align*}$

Now apply the chain rule to find these derivatives.
June 28th 2011, 08:00 AM
vidhi96

Re: differentiation applying two rules

This is where I get stuck- so I did prodcut rule and got as far as you have - now when you apply chain rule- which part do you aplly it to as I am thorughly confused.

Quote:

Originally Posted by Prove It

$\displaystyle \begin{align*}y &= e^{6x}(1 + 3x^2)^{\frac{1}{2}} \\ \frac{dy}{dx} &= e^{6x}\frac{d}{dx}\left[(1 + 3x^2)^{\frac{1}{2}}\right] + (1 + 3x^2)^{\frac{1}{2}}\frac{d}{dx}\left(e^{6x}\right) \end{align*}$

Now apply the chain rule to find these derivatives.
June 28th 2011, 08:26 AM
vidhi96

1 Attachment(s)

Re: differentiation applying two rules

Hi Please have a look at attachment of what I have done

Quote:

Originally Posted by Archie Meade

The Product Rule is

$(uv)'=u'(v)+v'(u)$

Hence you write your derivatives for u and v
and plug them into the formula.

This will be relatively easy if you can get those derivatives,
which require the Chain Rule.

How is this so far ?
June 28th 2011, 08:29 AM
vidhi96

1 Attachment(s)

Re: differentiation applying two rules

please have a look at what I have done and tell me if my understanding is correct or not Thanks
June 28th 2011, 08:36 AM
tom@ballooncalculus

Re: differentiation applying two rules

Wasn't able to open that, but...

Just in case a picture helps...

http://www.ballooncalculus.org/asy/diffProd/quot8.png

... where (key in spoiler) ...

Spoiler:

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside the right hand leg of the legs-uncrossed version of...

http://www.ballooncalculus.org/asy/prod.png

... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

Spoiler:

http://www.ballooncalculus.org/asy/diffProd/quot8a.png

Or, zooming in on the exponential also...

http://www.ballooncalculus.org/asy/diffProd/quot8b.png

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
June 28th 2011, 08:39 AM
vidhi96

Re: differentiation applying two rules

hey Guys is this the right answer
dy/dx=6e^6x 〖(1+3x^2)〗^(1/2)+e^6x 1/2 〖(1+3x^2)〗^(-1/2)
June 28th 2011, 08:44 AM
vidhi96

1 Attachment(s)

Re: differentiation applying two rules

ok heres a PDF hope you can open it- while I try and figure it out your diagram. I am not good at this cal but I must learn for my exam- so every help is much appriciated
June 28th 2011, 08:44 AM
tom@ballooncalculus

Re: differentiation applying two rules

Quote:

Originally Posted by vidhi96

hey Guys is this the right answer
dy/dx=6e^6x 〖(1+3x^2)〗^(1/2)+e^6x 1/2 〖(1+3x^2)〗^(-1/2)

What you're missing there is the 6x in the bottom-right corner of my diagrams. (Arising from the chain rule for differentiation, just as the 6 popped out from inside the exponential - you included that one, just do the same on the right, as per the diagram.)

PS the third (or fourth) balloon along has the sqrt tweaked into a fraction to prepare for simplifying.
June 28th 2011, 09:05 AM
vidhi96

1 Attachment(s)

Re: differentiation applying two rules

sorry I have self corrected myself and come to this final answer
June 28th 2011, 09:06 AM
vidhi96

Re: differentiation applying two rules

hey please see my correction thanks
June 28th 2011, 09:09 AM
Archie Meade

Re: differentiation applying two rules

Quote:

Originally Posted by vidhi96

ok heres a PDF hope you can open it- while I try and figure it out your diagram. I am not good at this cal but I must learn for my exam- so every help is much appriciated

You just need a bit more co-ordination.

On the attachment, you wrote

$\frac{dv}{dx}=\left(1+3x^2\right)^{\frac{1}{2}}$

This is in fact "v" written in "index form" in preparation for differentiation.

$v=\sqrt{1+3x^2}=\left(1+3x^2\right)^{\frac{1}{2}}$

You correctly differentiated $e^{6x}$
using the Chain Rule.

$e^{6x}=e^u$

$\frac{d}{dx}e^u=\frac{du}{dx}\frac{d}{du}e^u$

To differentiate "v"

$w=1+3x^2=f(x)$

$v=w^{\frac{1}{2}}=f(w)$

$\frac{dv}{dx}=\frac{dv}{dw}\frac{dw}{dx}$

Writing the stages in this way helps simplify the Chain Rule calculations.
June 28th 2011, 09:10 AM
vidhi96

Re: differentiation applying two rules

i am bit confident- I am going to work through another example and please guide me so I know I can do this. I will attach what I have done soon for you to look at thanks again
June 28th 2011, 09:15 AM
Archie Meade

Re: differentiation applying two rules

Quote:

Originally Posted by vidhi96

i am bit confident- I am going to work through another example and please guide me so I know I can do this. I will attach what I have done soon for you to look at thanks again

You will have it when you've mastered differentiating the "v" part.

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