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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
         ---------- =
             x³


  4. y = x²
         --------- =
           x+2x


   5 . f(x) = x-2
             ---------- =
                 x³



Please answer :)
i need Solution & perfectly answer tnx god bless mwaaahh...
  :*


Sagot :

1. f (x) = x²-4x
         ---------- =
           x+1

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

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

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

  4. y = x²
         --------- =
           x+2x
multiply 1/x to the numerator and denominator
you get:
=[tex]\frac{x}{1+2)[/tex]
=[tex]\frac[1}{3} (\frac{d}{dx} x)[/tex]
=1/3
   5 . f(x) = x-2
             ---------- =
                 x³
[tex]=(x-2)(x^(-3))
= x^(-3) - 3(x^(-4))(x-2)
=\frac{x-3x+6}{x^4}
=\frac{6-2x}{x^4}[/tex]