# differentiation applying two rules

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• Jun 28th 2011, 09:24 AM
vidhi96
Re: differentiation applying two rules
oh ! so is my answer incorrect ?
• Jun 28th 2011, 09:26 AM
vidhi96
Re: differentiation applying two rules
here is my second example - please see if my inital steps are correct- while I get my head around what I did wrong in first example- ta
• Jun 28th 2011, 09:29 AM
vidhi96
Re: differentiation applying two rules
what you mean?
• Jun 28th 2011, 09:34 AM
vidhi96
Re: differentiation applying two rules
Ok I read what you said and I tried again- you can't fail me for trying. Please have a look
• Jun 28th 2011, 09:43 AM
Re: differentiation applying two rules
Quote:

Originally Posted by vidhi96
what you mean?

For that last example, you overlooked squaring the entire denominator.

What I meant earlier is that you are comfortable with the situation
where the "function nesting" is one level deep, when applying the chain rule.

$\displaystyle e^{6x}$ is one level beyond $\displaystyle e^x$

q=6x

$\displaystyle \frac{d}{dx}e^{6x}=\frac{d}{dx}e^q=\frac{dq}{dx} \frac{d}{dq}e^q$

However...

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

has an extra stage.

Beginning with x, we have

$\displaystyle w=1+3x^2$

but then we take the square root of this

$\displaystyle v=w^{\frac{1}{2}}$

so the "nesting" is 2 levels deep, whereas it was only 1 level deep for $\displaystyle e^{6x}$

Hence

$\displaystyle \frac{d}{dx}\left(1+3x^2\right)^{\frac{1}{2}}= \frac{d}{dx}w^{\frac{1}{2}}=\frac{dw}{dx}\frac{d}{ dw}w^{\frac{1}{2}}$

$\displaystyle =\frac{dw}{dx}\left[\frac{1}{2}w^{-\frac{1}{2}}\right]$

How is that ?
• Jun 28th 2011, 09:48 AM
Re: differentiation applying two rules
Quote:

Originally Posted by vidhi96
Ok I read what you said and I tried again- you can't fail me for trying. Please have a look

On your line beginning "Therefore", you do not have the correct derivative for "v",
but you corrected yourself on the final line!
• Jun 28th 2011, 09:52 AM
vidhi96
Re: differentiation applying two rules
Quote:

Originally Posted by Prove It
\displaystyle \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.

Hey following some discussion with someone else and you- I have done some work- do you mind seeing what I have done is correct so I know I am ok to work with product and quotient rule both
• Jun 28th 2011, 09:53 AM
vidhi96
Re: differentiation applying two rules
so you are telling me I am finally right
• Jun 28th 2011, 09:55 AM
vidhi96
Re: differentiation applying two rules
Quote:

On your line beginning "Therefore", you do not have the correct derivative for "v",
but you corrected yourself on the final line!

I take it I finally understand how it works- Please say yes!!! I have also completed my own second eaxmple - heres my final version please look at it - I am typing so i can send it-therefore please excaue miner mistakes but please point to it
• Jun 28th 2011, 09:58 AM
Re: differentiation applying two rules
Quote:

Originally Posted by vidhi96
so you are telling me I am finally right

Yes, but do look at why you wrote the derivative of "v" incorrectly,
before plugging the correct derivative of v into the Product Rule formula.
For your Quotient Rule example, you still have not squared the denominator.
• Jun 28th 2011, 11:07 AM
vidhi96
Re: differentiation applying two rules
Quote: