G grapz Junior Member Joined Jan 13, 2007 Messages 80 Jan 29, 2007 #1 Suppose f is a function such that x[f(x)]^3 + x^2f(x) = 3 and f(2) = 1 find f' (2) i am not sure how to do this. do i just find the derivative of the first and plug it in? there is no answer so i'm really stuck.
Suppose f is a function such that x[f(x)]^3 + x^2f(x) = 3 and f(2) = 1 find f' (2) i am not sure how to do this. do i just find the derivative of the first and plug it in? there is no answer so i'm really stuck.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Jan 29, 2007 #2 Use the chain rule and product rule. You can sub in your values and solve for f'(2). \(\displaystyle \L\\x[f(x)]^{3}+x^{2}f(x)=3, \;\ f(2)=1\) \(\displaystyle \L\\x\underbrace{(3[f(x)]^{2}f'(x))}_{\text{chain rule}}+[f(x)]^{3}+x^{2}f'(x)+f(x)2x\)
Use the chain rule and product rule. You can sub in your values and solve for f'(2). \(\displaystyle \L\\x[f(x)]^{3}+x^{2}f(x)=3, \;\ f(2)=1\) \(\displaystyle \L\\x\underbrace{(3[f(x)]^{2}f'(x))}_{\text{chain rule}}+[f(x)]^{3}+x^{2}f'(x)+f(x)2x\)
