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# Calculus help... On derivatives. Relatively simple, just confused?

Find the derivative--

f(x) = x [ 1 - (4 / x + 3) ]

Thanks!

### 4 Answers

- BleakskyLv 61 decade agoFavorite Answer
f(x) = x[1 - (4 / x + 3)]

Rewrite the function to make it a bit easier to find the derivative.

x(1 - (4/x) - 3)

x(-2 - (4/x))

f (x) = -2x - 4

f ' (x) = -2

- 1 decade ago
Use the Chain Rule.

Let g(x) = x, and h(x) = 1 - (4/x + 3).

The Chain Rule states f(x) = g(x)h(x) -> f'(x) = g'(x)h(x) + g(x)h'(x).

So, f'(x) = 1- (4/x + 3) + x[8/(x^2)].

Simplification yields 1-4/x - 3 + 4/x = -2

- 1 decade ago
you could use product rule: der of the 1st times the 2nd+der of the 2nd times the first

but since the first is only a monomial (x) multiply the x through then use power rule

the second part uses quotient rule:der top times the bottom -der bottom times the top all over the bottom squared

f(x)=x- 4x/(x+3)

f '(x)=1- (4(x+30)-1(4x))

(x+3)^2

so f '(x)=1- 4x+120-4x =1- 120

(x+3)^2 (x+3)^2

just algebra after that

- mohanrao dLv 71 decade ago
f(x) = x [ 1 - (4 / x + 3) ]

f ' (x) = x [ 0 - ( - 4 / (x + 3)^2 ] + [ 1 - (4 / x + 3) ]

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

= [ 4x + x^2 + 2x - 3 ] / (x + 3)^2

= (x^2 + 6x - 3 ) / (x + 3)^2