# Math Help - differentiation applying two rules

1. ## Re: differentiation applying two rules

oh ! so is my answer incorrect ?

2. ## 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

3. ## Re: differentiation applying two rules

what you mean?

4. ## 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

5. ## Re: differentiation applying two rules

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.

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

q=6x

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

However...

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

has an extra stage.

Beginning with x, we have

$w=1+3x^2$

but then we take the square root of this

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

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

Hence

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

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

How is that ?

6. ## Re: differentiation applying two rules

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!

7. ## Re: differentiation applying two rules

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.
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

8. ## Re: differentiation applying two rules

so you are telling me I am finally right

9. ## Re: differentiation applying two rules

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

10. ## Re: differentiation applying two rules

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.

11. ## Re: differentiation applying two rules

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.
Ok I will do squaring and see if I can simplify in any way. However I take it rest of is ok. Is that correct? Also I know why I made mistake in "v" because I wasn't careful but since you pointed it out I think I will be extra carful in writing my steps out. Just away from the computer for next hour so when get back will send you my correction. I must say you have given me that confidence I thought I'll never have . Sincere thank you

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