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Homework Help: Differentiation - slopes

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the slope of the curve [itex]y=\frac{x}{3x+2} [/itex] at the point x = -2.

    2. Relevant equations
    \lim_{h \rightarrow 0}\frac{f(x_{0} + h) - f(x_{0})}{h} = m

    3. The attempt at a solution
    If x = -2, then y = 1/2. I'm not sure what to do from here.

    This is the first step, but I don't get how you obtain this:
    \lim_{h \rightarrow 0}\frac{\frac{-2+h}{3(-2+h)+2}-\frac{1}{2}}{h}

    I just need an explanation of what we're doing here and why?

    Thanks in advance for the help!
    Last edited: Oct 25, 2009
  2. jcsd
  3. Oct 25, 2009 #2


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    This would be very difficult to get from the definition of a limit. You probably have something called the quotient rule that will help greatly
  4. Oct 25, 2009 #3
    The differentiation rules aren't introduced for another one and a half chapters in my text book. This part is introducing a definition to the slope of a curve. It says:

    The slope of a curve C at a point P is the slope of the tangent line to C at P if such a tangent line exists. In particular, the slope of the graph of y = f(x) at the point x0 is
    \lim_{h \rightarrow 0}\frac{f(x_{0} + h) - f(x_{0})}{h} = m[/tex]

    And after this definition, it gives the example that I posted. I have the answer, just don't understand any of it :(
  5. Oct 25, 2009 #4


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    This can be done with some algebraic manipulation. Let your first term in the numerator be [tex]\frac{-2+h}{3h-4}[/tex] and get the second term of 1/2 into that same form so that you can take [tex]3h-4[/tex] under the denominator of the whole and go from there.
  6. Oct 25, 2009 #5


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    [tex]f(x)= \frac{x}{3x+ 2}[/tex]
    so [tex]f(x_0+h)= \frac{x_0+h}{3(x_0+h)+ 2}= \frac{x_0+h}{3x_0+3h+2}[/tex]

    [tex]f(x_0+h)- f(x_0)= \frac{x_0+h}{3(x_0+h)+ 2}= \frac{x_0+h}{3x_0+3h+2}- \frac{x}{3x+ 2}[/tex]

    The "common denominator" is [itex](3x_0+ 2)(3x_0+ 3h+ 2)[/itex]. Multiplying numerator and denominator of the first fraction by [itex]3x_0+ 2[/itex] and the numerator and denominator of the second fraction by [itex]3x_0+ 3h+ 2[/itex],

    [tex]\frac{(x_0+h)(3x_0+ 2)}{(3x_0+ 3h+ 2)(3x_0+ 2)}- \frac{(x_0)(3x_0+ 3h+ 2)}{(3x_0+ 3h+ 2)(3x_0+ 2)}[/tex]

    Multiply out the products in the numerators. You can leave the denominators as they are:
    [tex]\frac{3x_0^2+ 2x_0+ 3hx_0+ 2h}{(3x_0+ 3h+ 2)(3x_0+ 2)}- \frac{3x_0^2+ 3hx_0+ 2x_0}{(3x_0+ 3h+ 2)(3x_0+ 2)}[/tex]

    Now you see that the "[itex]3x_0^2[/itex]", "[itex]3hx_0[/itex]", and "[itex]2x_0[/itex]" terms cancel leaving
    [tex]\frac{2h}{(3x_0+ 3h+ 2)(3x_0+ 2)}[/tex]

    That is [itex]f(x_0+ h)- f(x_0)[/itex]. To form the "difference quotient" divide by h:
    [tex]\frac{f(x_0+ h)- f(x_0)}{h}= \frac{2h}{h(3x_0+ 3h+ 2)(3x_0+ 2)}= \frac{2}{(3x_0+ 3h+ 2)(3x_0+ 2)}[/tex].

    Finally, take the limit as h goes to 0. Since setting h to 0 does not make the denominator 0, you can do that simply by setting h= 0.
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