
#1
Oct1507, 12:02 AM

P: 52

Question:
If [tex]f(x) + x^{2}[f(x)]^{3} = 10[/tex] and f(1) = 2, find f '(1). Attempt: 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 :) 



#2
Oct1507, 12:05 AM

P: 372

Take the derivative of both sides of the equation.




#3
Oct1507, 12:12 AM

P: 8

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).




#4
Oct1507, 12:39 AM

P: 52

Derivative of a Function
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? So... 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 so.. [tex]f'(1) + (1)^{2}(3(2)^{2}*f'(1)) + (2)^{3}(2(1)) = 0[/tex] then [tex]f'(1) + 12*f'(1)) + 16 = 0[/tex] which means [tex](13)f'(1)) = 16[/tex] making [tex]f'(1) = 16/13[/tex]??? Does that make sense? Is it wrong? i have a feeling it is :/ 



#5
Oct1507, 12:55 AM

P: 492

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). 



#6
Oct1507, 12:07 PM

P: 87

f(x)+x^(2)*f(x)^(3)//use product rule for the second term along with the chain rule
f'(x)+x^(2)*3f(x)^(2)f'(x)+2xf(x)^(3) f'(x)[1+3x^(2)]=2xf(x)^(3)//moved 2xf(x)^(3) to other side and factored a f'(x) f'(x)=[2xf(x)^(3)]/[1+3x^(2)] f'(1)=[2(1)f(1)^3]/[1+3(1)^(2)} f'(1)=16/13 OWNED!!! 


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