CALCULUS :

Direction: Find the derivatives of the following functions using any of the two methods


1. f (x) = x²-4x
---------- =
x+1


2. f (x) = 3x
-------- =
x +4


3. f(x)= x+1
---------- =



4. y = x²
--------- =
x+2x


5 . f(x) = x-2
---------- =




Please answer :)
i need Solution & perfectly answer tnx god bless mwaaahh...
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1
i don't know d answer

Answers

2014-01-21T17:13:03+08:00
1. f (x) = x²-4x
         ---------- =
           x+1

= (x²-4x)(x+1)^-1
remember that (f)(g) = f'g+g'f
=\frac{2x-4}{x+1} - \frac{x^2 - 4x}{(x+1)^2}
= \frac{x^2 +2x - 4}{(x+1)^2}

  2. f (x) =  3x
             -------- =
              x +4
= 3x(x+4)^-1
=3x(-(x+4)^2) + 3((x+4)^-1)
=12/(x+4)^2

   3. f(x)= x+1
         ---------- =
             x³
=(x+10)((x)^(-3))
=x^(-3) - 3(x^(-4))(x+1)
=\frac{-2x+3}{x^4}

  4. y = x²
         --------- =
           x+2x
multiply 1/x to the numerator and denominator
you get:
=\frac{x}{1+2)
=\frac[1}{3} (\frac{d}{dx} x)
=1/3
   5 . f(x) = x-2
             ---------- =
                 x³
=(x-2)(x^(-3)) = x^(-3) - 3(x^(-4))(x-2) =\frac{x-3x+6}{x^4} =\frac{6-2x}{x^4}
0
sorry in number 4, there might've been a problem with how i wrote the equation....anyway it's supposed to be x/(2+1). Since you can remove the constant (1/3), you just need to get the derivative of x which is 1. So (1/3)(1) = 1/3