Physics Forums

Physics Forums (
-   Calculus & Beyond Homework (
-   -   Derivative of a Function (

momogiri Oct15-07 12:02 AM

Derivative of a Function
If [tex]f(x) + x^{2}[f(x)]^{3} = 10[/tex] and f(1) = 2, find f '(1).

I wish I could say I tried, but I don't know how to approach this problem..
All I did was double check the formula worked by inputting 2 for f(x) and 1 for x

Can someone tell me how to start this? And I'll go from there :)

ZioX Oct15-07 12:05 AM

Take the derivative of both sides of the equation.

NonAbelian Oct15-07 12:12 AM

And don't forget to use the chain rule on the second part. You'll end up with an equation that involves both f(x) and f'(x). At this point, sub in x=1 and f(1)=2 and solve for f'(x).

momogiri Oct15-07 12:39 AM

Ok, so what I've done is:
take derivative of both sides
original = [tex]f(x) + x^{2}(f(x))^{3} = 10[/tex]
so d/dx of f(x) is f'(x)
now [tex]x^{2}(f(x))^{3}[/tex]'s derivative I'm a bit unsure of..

What I did was use the product rule, so..
[tex]x^{2}((f(x))^{3})' + (f(x))^{3}(x^{2})'[/tex]
and [tex](f(x)^{3})' = 3(f(x))^{2}*f'(x)[/tex] right?
it's [tex]x^{2}(3(f(x))^{2}*f'(x)) + f(x)^{3}(2x)[/tex]
Then that means the whole equation becomes
[tex]f'(x) + x^{2}(3(f(x))^{2}*f'(x)) + f(x)^{3}(2x) = 0[/tex]
In which I plugged in the numbers
[tex]f'(1) + (1)^{2}(3(2)^{2}*f'(1)) + (2)^{3}(2(1)) = 0[/tex]
[tex]f'(1) + 12*f'(1)) + 16 = 0[/tex]
which means
[tex](13)f'(1)) = -16[/tex]
[tex]f'(1) = -16/13[/tex]???
Does that make sense? Is it wrong? i have a feeling it is :/

bob1182006 Oct15-07 12:55 AM

looks right.

if you wanted to not deal with f(x) you could change f(x)=y and do implicit differentiation like you've been doing before and solve for y' and then change y to f(x).

eyehategod Oct15-07 12:07 PM

f(x)+x^(2)*f(x)^(3)//use product rule for the second term along with the chain rule
f'(x)[1+3x^(2)]=-2xf(x)^(3)//moved 2xf(x)^(3) to other side and factored a f'(x)

All times are GMT -5. The time now is 02:05 PM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums