## Derivative of a Function

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 :)

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 Take the derivative of both sides of the equation.
 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).

## 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 :/

 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).
 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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