# Derivative of a Function

by momogiri
Tags: derivative, function
 P: 52 Question: If $$f(x) + x^{2}[f(x)]^{3} = 10$$ 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 :)
 P: 52 Derivative of a Function Ok, so what I've done is: take derivative of both sides original = $$f(x) + x^{2}(f(x))^{3} = 10$$ so d/dx of f(x) is f'(x) now $$x^{2}(f(x))^{3}$$'s derivative I'm a bit unsure of.. What I did was use the product rule, so.. $$x^{2}((f(x))^{3})' + (f(x))^{3}(x^{2})'$$ and $$(f(x)^{3})' = 3(f(x))^{2}*f'(x)$$ right? So... it's $$x^{2}(3(f(x))^{2}*f'(x)) + f(x)^{3}(2x)$$ Then that means the whole equation becomes $$f'(x) + x^{2}(3(f(x))^{2}*f'(x)) + f(x)^{3}(2x) = 0$$ In which I plugged in the numbers so.. $$f'(1) + (1)^{2}(3(2)^{2}*f'(1)) + (2)^{3}(2(1)) = 0$$ then $$f'(1) + 12*f'(1)) + 16 = 0$$ which means $$(13)f'(1)) = -16$$ making $$f'(1) = -16/13$$??? Does that make sense? Is it wrong? i have a feeling it is :/