Question

F(x)+x4[f(x)]3=2005f(x)+x4[f(x)]3=2005 and f(2)=5f(2)=5, find f′(2)

PLEASE HELP
We went over this in class but she only showed us how to get it to dy/dx and not how to actually get a number I REALLY need help on this!
Calculus is horrible

Answer 1

`f(x)+x^4f(x)^3=2005`

Differentiating both sides yields


`(\mathrm d)/(\mathrm dx)f(x)+(\mathrm d)/(\mathrm dx)(x^4f(x)^3)=(\mathrm d)/(\mathrm dx)2005`

The right hand side is a constant, so the derivative is just 0, and
`(\mathrm d)/(\mathrm dx)f(x)=f'(x)`. The remaining term requires using the product rule:


`(\mathrm d)/(\mathrm dx)(x^4f(x)^3)=f(x)^3(\mathrm d)/(\mathrm dx)x^4+x^4(\mathrm d)/(\mathrm dx)f(x)^3`

Using the power rule for the first term here and the power/chain rules for the second gives


`(\mathrm d)/(\mathrm dx)(x^4f(x)^3)=4x^3f(x)^3+3x^4f(x)^2(\mathrm d)/(\mathrm dx)f(x)`

`(\mathrm d)/(\mathrm dx)(x^4f(x)^3)=4x^3f(x)^3+3x^4f(x)^2f'(x)`

Putting everything together, you have


`f'(x)+4x^3f(x)^3+3x^4f(x)^2f'(x)=0`

Collecting the terms with
`f'(x)` gives


`(1+3x^4f(x)^2)f'(x)+4x^3f(x)^3=0`

`(1+3x^4f(x)^2)f'(x)=-4x^3f(x)^3`

`f'(x)=-(4x^3f(x)^3)/(1+3x^4f(x)^2)`

Now, finding the value of the derivative at
`x=2` is just a matter of plugging in
`x=2`:


`f'(2)=-(4*2^3f(2)^3)/(1+3*2^4f(2)^2)`

We know
`f(2)=5`, so we get


`f'(2)=-(4*2^3*5^3)/(1+3*2^4*5^2)`

`f'(2)=-(4000)/(1201)`