f(x) + x^2[f(x)]^3 = 10, f(1)=2; f'(1)=?
I tried working through this best as I know how, but didn't get the answer my book has... Can someone explain to me how to solve this?
Any and all help is greatly appreciated!

*Answer in my book: -16/13

Question

f(x) + x^2[f(x)]^3 = 10, f(1)=2; f'(1)=?
I tried working through this best as I know how, but didn't get the answer my book has... Can someone explain to me how to solve this? 
Any and all help is greatly appreciated!

*Answer in my book: -16/13

amonoconnor · Answer

Here's what I did:
$$d/dx[f(x) + x^2 [f(x)]^3] = $$
$$(1) + [2x(f(x)^3] + x^2(3f(x)^2)(1)$$
$$1+2x(f(x)^3) + 3x^2(f(x)^2)$$

amonoconnor · Answer

Then I plugged in 1 for x, and 2 for f(x)... and got 29.

zepdrix · Answer

Hmm those 1's don't look right...$$\large\rm \frac{d}{dx}f(x)=f'(x)$$$$\large\rm \frac{d}{dx}f(x)\ne 1$$

amonoconnor · Answer

But what do I plus in for f'(x) then, numerically?

zepdrix · Answer

$$\large\rm \frac{d}{dx}\left(f+x^2f^3\right)=f'+2xf^3+x^2(3f^2)f'=0$$Your product rule looks good though :)

amonoconnor · Answer

I was evaluating f(x) like a "y".

zepdrix · Answer

Well you should have f'(x) showing up SOMEWHERE,
seeing is that is what the problem is asking you to solve for, ya? :)

freckles · Answer

if you was you would have been writing d(y)/dx as dy/dx or y' or f'

zepdrix · Answer

$$\large\rm \frac{d}{dx}f(x)=\frac{d~f(x)}{dx}=f'(x)$$
Ya it's the same as the dy/dx that you've dealt with in previous problems :)

zepdrix · Answer

Did you get that computer charged back up? XD hah

amonoconnor · Answer

So I write any f(x)'s as: "1*dy/dx"?  And then pull all dy/dx's to one side of the equals sign?

zepdrix · Answer

Yes. And then do that factor business.

zepdrix · Answer

Well you can write any (d/dx)f(x)'s as what you wrote.

amonoconnor · Answer

And yep, I'm in the comforts of home now;)

zepdrix · Answer

noiceee :U

johnweldon1993 · Answer

You're actually very close if you got 29, that's what I was getting until I noticed the dumb mistake I made XD

amonoconnor · Answer

Ahh! Help!!

johnweldon1993 · Answer

Show us what you have when you complete taking the derivative :D

amonoconnor · Answer

Will do!

johnweldon1993 · Answer

*You'll probably have it figured out before you even need to type it out :P

zepdrix · Answer

$$\large\rm f+x^2f^3=10$$$$\large\rm f'+2xf^3+3x^2f^2f'=0$$Ya, were you able to solve for f' at this point? :)
You can ignore my notation if you don't like it.
I just hate rewriting f'(x) or dy/dx over and over.

johnweldon1993 · Answer

I agree^ lol

amonoconnor · Answer

Yep, I'm on board, and have...

amonoconnor · Answer

$$dy/dx = \frac{2x(f(x)^3}{3x^2(f(x)^2)+1}$$

amonoconnor · Answer

Again, sorry about these stupid-looking fractions... They USED to be normal :p

johnweldon1993 · Answer

That took me a second I was like "the hell did he write?" haha but hmm, I see 1 mistake...hmm, plug in f(1) = 2 everywhere and see what you get :P lets make this an experiment XD

amonoconnor · Answer

Ah... that's where that missing negative I was just looking for comes from... touche. I missed it! I hate those damn signs...

johnweldon1993 · Answer

Lol well there ya go :D

amonoconnor · Answer

And when I plug in that stuff, I get -16/13!!! :D

zepdrix · Answer

woo

amonoconnor · Answer

Thank you guys, again lols. You rock;) AND save my retrice repeatedly, when I just don't understand processes. I'm a huge like, need to understand what I'm doing, not just follow a formula guy, so having people to bounce ideas and problems off of makes homework... get done lols. My professor sucks, and is so hard to engage...

amonoconnor · Answer

lols, so OS turned a$$ into *retirce

johnweldon1993 · Answer

Understandable, that's what most of the people I tutor are like as well

But yeah I wish some of my professors were more engaging as well, oh well, that's what we're here for :D