# Confunsing Chain Rules and more

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Mar 14th 2009, 06:44 PM
sk8erboyla2004
Confunsing Chain Rules and more
I need a bit of help with my assignment it due monday but I am in utter confusement.

Here is my first problem I need help making I did right

someone please check this derivative for me

the function is
$\displaystyle f(x)=e^{2x}ln(10x^2+3)$

$\displaystyle f\prime(x)=(2e^{2x})[ln(10x^{2}+3)]+(e^{2x})(\frac{1}{10x^2+3})$
$\displaystyle f\prime(x)=e^{2x}[2(ln(10x^2+3)]+(\frac{1}{10x^2+3})$

?

Also how would I begin to simplify something like this take the derivativea

$\displaystyle -\frac{x^2+x^3-\pi}{x^-3}$?
• Mar 14th 2009, 06:55 PM
Prove It
Quote:

Originally Posted by sk8erboyla2004
I need a bit of help with my assignment it due monday but I am in utter confusement.

Here is my first problem I need help making I did right

someone please check this derivative for me

the function is
$\displaystyle f(x)=e^{2x}ln(10x^2+3)$

$\displaystyle f\prime(x)=(2e^{2x})[ln(10x^{2}+3)]+(e^{2x})(\frac{1}{10x^2+3})$
$\displaystyle f\prime(x)=e^{2x}[2(ln(10x^2+3)]+(\frac{1}{10x^2+3})$

?

Also how would I begin to simplify something like this take the derivativea

$\displaystyle -\frac{x^2+x^3-\pi}{x^-3}$?

If you have combinations of product, chain or quotient rules, it helps to take things one step at a time.

So for the first...

$\displaystyle f = e^{2x}\ln{(10x^2 + 3)}$

Use the product rule

$\displaystyle \frac{df}{dx} = e^2x\frac{d}{dx}[\ln{(10x^2 + 3)}] + \ln{(10x^2 + 3)}\frac{d}{dx}(e^{2x})$.

Now use the chain rule to work out these easier derivatives.

$\displaystyle \frac{d}{dx}(e^{2x})$ is easy, it's just $\displaystyle 2e^{2x}$.

If $\displaystyle y = \ln{(10x^2 + 3)}$ let $\displaystyle u = 10x^2 + 3$ so that $\displaystyle y = \ln{u}$.

$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du}$.

$\displaystyle \frac{du}{dx} = 20x$

$\displaystyle \frac{dy}{du} = \frac{1}{u} = \frac{1}{10x^2 + 3}$.

So $\displaystyle \frac{dy}{dx} = \frac{20x}{10x^2 + 3}$.

Therefore

$\displaystyle \frac{df}{dx} = \frac{20xe^{2x}}{10x^2 + 3} + 2e^{2x}\ln{(10x^2 + 3)}$.
• Mar 14th 2009, 07:01 PM
Prove It
Quote:

Originally Posted by sk8erboyla2004
I need a bit of help with my assignment it due monday but I am in utter confusement.

Here is my first problem I need help making I did right

someone please check this derivative for me

the function is
$\displaystyle f(x)=e^{2x}ln(10x^2+3)$

$\displaystyle f\prime(x)=(2e^{2x})[ln(10x^{2}+3)]+(e^{2x})(\frac{1}{10x^2+3})$
$\displaystyle f\prime(x)=e^{2x}[2(ln(10x^2+3)]+(\frac{1}{10x^2+3})$

?

Also how would I begin to simplify something like this take the derivativea

$\displaystyle -\frac{x^2+x^3-\pi}{x^-3}$?

$\displaystyle f = -\frac{x^2+x^3-\pi}{x^{-3}}$ is a quotient, so use the quotient rule.

$\displaystyle \frac{df}{dx} = -\left[\frac{x^{-3}\frac{d}{dx}(x^2 + x^3 - \pi) - (x^2 + x^3 - \pi)\frac{d}{dx}(x^{-3})}{(x^{-3})^2}\right]$

$\displaystyle = -\left[\frac{x^{-3}(2x + 3x^2) + 3x^{-4}(x^2 + x^3 - \pi)}{x^{-6}}\right]$

$\displaystyle = -[x^3(2x + 3x^2) + 3x^2(x^2 + x^3 - \pi)]$.
• Mar 14th 2009, 07:20 PM
sk8erboyla2004
Quote:

Originally Posted by Prove It
If you have combinations of product, chain or quotient rules, it helps to take things one step at a time.

So for the first...

$\displaystyle f = e^{2x}\ln{(10x^2 + 3)}$

Use the product rule

$\displaystyle \frac{df}{dx} = e^2x\frac{d}{dx}[\ln{(10x^2 + 3)}] + \ln{(10x^2 + 3)}\frac{d}{dx}(e^{2x})$.

Now use the chain rule to work out these easier derivatives.

$\displaystyle \frac{d}{dx}(e^{2x})$ is easy, it's just $\displaystyle 2e^{2x}$.

If $\displaystyle y = \ln{(10x^2 + 3)}$ let $\displaystyle u = 10x^2 + 3$ so that $\displaystyle y = \ln{u}$.

$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du}$.

$\displaystyle \frac{du}{dx} = 20x$

$\displaystyle \frac{dy}{du} = \frac{1}{u} = \frac{1}{10x^2 + 3}$.

So $\displaystyle \frac{dy}{dx} = \frac{20x}{10x^2 + 3}$.

Therefore

$\displaystyle \frac{df}{dx} = \frac{20xe^{2x}}{10x^2 + 3} + 2e^{2x}\ln{(10x^2 + 3)}$.

but isnt the derivative of the natural log 1/x? so where does that come into play if it does

and thanks!

nvm I think I understood that part man I am sure dumbfounded as when to use the cahin rule
• Mar 14th 2009, 07:27 PM
Prove It
Quote:

Originally Posted by sk8erboyla2004
but isnt the derivative of the natural log 1/x? so where does that come into play if it does

and thanks!

Yes, the derivative of a natural logarithm of the form

$\displaystyle y = \ln{x}$ is $\displaystyle \frac{1}{x}$.

But if you have a natural logarithm of a FUNCTION of x, you have to use the chain rule.

I've shown it to you step by step already...
• Mar 14th 2009, 07:38 PM
sk8erboyla2004
Yes ive seen but I have no clue how to implement this chain rule properly
• Mar 14th 2009, 07:40 PM
Prove It
• Mar 14th 2009, 07:49 PM
sk8erboyla2004
Yes, I understand and I can clearly see in the example of that chain rule is

$\displaystyle f\prime(g(x))*g\prime(x)$

and that

$\displaystyle f\prime(g(x)=\frac{1}{10x^2+3}$

and

$\displaystyle g\prime(x)=20x$

What I rather meant to say is I am unclear of when to use the chain rule and the extent of using it for more than two function as per say

Thanks though!

check my answer for this one as well

$\displaystyle y=\frac{e^{\pi}}{\pi^{e}}$

$\displaystyle y\prime=(e^{\pi})(\pi^{e}+e\pi^{e-1})$
• Mar 14th 2009, 07:59 PM
Prove It
Quote:

Originally Posted by sk8erboyla2004
Yes, I understand and I can clearly see in the example of that chain rule is

$\displaystyle f\prime(g(x))*g\prime(x)$

and that

$\displaystyle f\prime(g(x)=\frac{1}{10x^2+3}$

and

$\displaystyle g\prime(x)=20x$

What I rather meant to say is I am unclear of when to use the chain rule and the extent of using it for more than two function as per say

Thanks though!

check my answer for this one as well

$\displaystyle y=\frac{e^{\pi}}{\pi^{e}}$

$\displaystyle y\prime=(e^{\pi})(\pi^{e}+e\pi^{e-1})$

Yes it can be difficult. You just need practice.

Um, for your next one, $\displaystyle \frac{e^\pi}{\pi^e}$ is a constant.

So it's derivative is 0.
• Mar 14th 2009, 08:17 PM
sk8erboyla2004
Quote:

Originally Posted by Prove It
Yes it can be difficult. You just need practice.

Um, for your next one, $\displaystyle \frac{e^\pi}{\pi^e}$ is a constant.

So it's derivative is 0.

Darn I need mad practice, ugh.

Thanks also check this one it has far as i got and im unsure of how to future simply it bc i maybe be wrong

$\displaystyle h(x)=\frac{(2x^2-4x^6)^4}{2+2x^2}$

$\displaystyle h\prime(x)=\frac{4(2x^2-4x^6)^3(4x-24x^5)(2+2x^2)-(4x)(2x^2-4x^6)^4}{(2+2x^2)^2}$

$\displaystyle h\prime(x)=\frac{4(2x^2-4x^6)^3(4x-24x^5)-(4x)(2x^2-4x^6)^4}{(2+2x^2)}$

I am 75% sure I think I got this one correct as well

$\displaystyle f(x)=ln(2^x-3^{\frac{1}{2}}$
$\displaystyle f\prime(x)=\frac{(ln2(2^x)-\frac{3}{2\sqrt{x}}}{(2^x-3\sqrt{x})}$

This also

$\displaystyle y=\sqrt{3e^x-x^\pi}$

$\displaystyle y\prime=\frac{1}{2}(3e^x-x^\pi) * (3e^x-\pi^{\pi-1})$

Sorry
• Mar 14th 2009, 08:26 PM
Prove It
Quote:

Originally Posted by sk8erboyla2004
Darn I need mad practice, ugh.

Thanks also check this one it has far as i got and im unsure of how to future simply it bc i maybe be wrong

$\displaystyle h(x)=\frac{(2x^2-4x^6)^4}{2+2x^2}$

$\displaystyle h\prime(x)=\frac{4(2x^2-4x^6)^3(4x-24x^5)(2+2x^2)-(4x)(2x^2-4x^6)^4}{(2+2x^2)^2}$

$\displaystyle h\prime(x)=\frac{4(2x^2-4x^6)^3(4x-24x^5)-(4x)(2x^2-4x^6)^4}{(2+2x^2)}$

Like I say, it's always easier to break it into smaller parts.

$\displaystyle h = \frac{(2x^2 - 4x^6)^4}{2 + 2x^2}$

$\displaystyle \frac{dh}{dx} = \frac{(2 + 2x)\frac{d}{dx}[(2x^2 - 4x^6)^4] - (2x^2 - 4x^6)^4\frac{d}{dx}(2 + 2x^2)}{(2 + 2x^2)^2}$

For the difficult looking derivative, use the chain rule.

$\displaystyle y = (2x^2 - 4x^6)^4$

Let $\displaystyle u = 2x^2 - 4x^6$ so that $\displaystyle y = u^4$

$\displaystyle \frac{du}{dx} = 4x - 24x^5, \frac{dy}{du} = 4u^3 = 4(2x^2 - 4x^6)^3$.

So $\displaystyle \frac{dy}{dx} = 4(4x - 24x^5)(2x^2 - 4x^6)^3$.

Therefore

$\displaystyle \frac{dh}{dx} = \frac{4(2 + 2x)(4x - 24x^5)(2x^2 - 4x^6)^3 - 4x(2x^2- 4x^6)^4}{(2 + 2x)^2}$.
• Mar 14th 2009, 09:43 PM
sk8erboyla2004
I had to make a separate post for this one

$\displaystyle y=(\frac{x^3+2x}{e^{x^2}})^3$

$\displaystyle y\prime=3(\frac{x^3+2x}{e^{x^2}})[(3x^2+2)(e^{x^2}-(x^3+2)(e^{x^2})(2x)]$

if this is right how can i further simplify it?
• Mar 14th 2009, 09:44 PM
sk8erboyla2004
Quote:

Originally Posted by Prove It
Like I say, it's always easier to break it into smaller parts.

$\displaystyle h = \frac{(2x^2 - 4x^6)^4}{2 + 2x^2}$

$\displaystyle \frac{dh}{dx} = \frac{(2 + 2x)\frac{d}{dx}[(2x^2 - 4x^6)^4] - (2x^2 - 4x^6)^4\frac{d}{dx}(2 + 2x^2)}{(2 + 2x^2)^2}$

For the difficult looking derivative, use the chain rule.

$\displaystyle y = (2x^2 - 4x^6)^4$

Let $\displaystyle u = 2x^2 - 4x^6$ so that $\displaystyle y = u^4$

$\displaystyle \frac{du}{dx} = 4x - 24x^5, \frac{dy}{du} = 4u^3 = 4(2x^2 - 4x^6)^3$.

So $\displaystyle \frac{dy}{dx} = 4(4x - 24x^5)(2x^2 - 4x^6)^3$.

Therefore

$\displaystyle \frac{dh}{dx} = \frac{4(2 + 2x)(4x - 24x^5)(2x^2 - 4x^6)^3 - 4x(2x^2- 4x^6)^4}{(2 + 2x)^2}$.

SO I was somewhere close?
• Mar 14th 2009, 10:26 PM
Prove It
Quote:

Originally Posted by sk8erboyla2004
I had to make a separate post for this one

$\displaystyle y=(\frac{x^3+2x}{e^{x^2}})^3$

$\displaystyle y\prime=3(\frac{x^3+2x}{e^{x^2}})[(3x^2+2)(e^{x^2}-(x^3+2)(e^{x^2})(2x)]$

if this is right how can i further simplify it?

The easiest way to make mistakes is to skip steps.

Write down EVERYTHING.

$\displaystyle y = \left(\frac{x^3 + 2x}{e^{x^2}}\right)^3$.

You should be able to see that it is a quotient taken to a power. So use the chain rule first (as the "final" thing done was to take it to a power...)

So Let $\displaystyle u = \frac{x^3 + 2x}{e^{x^2}}$ so that $\displaystyle y = u^3$.

$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du}$.

$\displaystyle \frac{dy}{du}$ is easy, it's

$\displaystyle 3u^2 = 3\left(\frac{x^3 + 2x}{e^{x^2}}\right)^2$.

$\displaystyle \frac{du}{dx}$ is harder, as it's a quotient.

$\displaystyle \frac{du}{dx} = \frac{e^{x^2}\frac{d}{dx}(x^3 + 2x) - (x^3 + 2x)\frac{d}{dx}(e^{x^2})}{(e^{x^2})^2}$

$\displaystyle = \frac{e^{x^2}(3x^2 + 2) - 2xe^{x^2}(x^3 + 2x)}{e^{2x^2}}$.

Therefore

$\displaystyle \frac{dy}{dx} = \frac{3[e^{x^2}(3x^2 + 2) - 2xe^{x^2}(x^3 + 2x)]}{e^{2x^2}}\left(\frac{x^3 + 2x}{e^{x^2}}\right)^2$.
• Mar 14th 2009, 10:38 PM
sk8erboyla2004
I really appreciate this bro, with each problem Im getting better with the last one you just posted i forgot to add the /g(x)^2 to the bottom

Thanks Ill have many more to come lol!

like this one

$\displaystyle r(p)=tan^3(4p)$

$\displaystyle r\prime(p)=\frac{4}{cos^3(4p)}$
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last