# differentiation applying two rules

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• Jun 28th 2011, 06:50 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
• Jun 28th 2011, 06:55 AM
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 ?
• Jun 28th 2011, 06:58 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.
• Jun 28th 2011, 09:00 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.

• Jun 28th 2011, 09:26 AM
vidhi96
Re: differentiation applying two rules
Hi Please have a look at attachment of what I have done

Quote:

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 ?

• Jun 28th 2011, 09:29 AM
vidhi96
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
• Jun 28th 2011, 09:36 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.

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• Jun 28th 2011, 09:39 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)
• Jun 28th 2011, 09:44 AM
vidhi96
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
• Jun 28th 2011, 09:44 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.
• Jun 28th 2011, 10:05 AM
vidhi96
Re: differentiation applying two rules
sorry I have self corrected myself and come to this final answer
• Jun 28th 2011, 10:06 AM
vidhi96
Re: differentiation applying two rules
hey please see my correction thanks
• Jun 28th 2011, 10:09 AM
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.
• Jun 28th 2011, 10:10 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
• Jun 28th 2011, 10:15 AM