The Derivative by Definition

Key Questions

  • First you have to calculate the derivative of the function.

    #f(x)=x^3#

    #f'(x)=3x^2#

    Then if we want to find the derivative of #f(x)# when #x=4# then we substitute that value into #f'(x)#.

    #f'(4)=3(4)^2=3*16=48#

  • The formal definition of derivative of a function #y=f(x)# is:
    #y'=lim_(Deltax->0)(f(x+Deltax)-f(x))/(Deltax)#

    The meaning of this is best understood observing the following diagram:

    enter image source here

    The secant PQ represents the mean rate of change #(Deltay)/(Deltax)# of your function in the interval between #x# and #x+Deltax#.

    If you want the rate of change, say, at P you "move" point Q (and the secant with it) to meet point P as in:
    enter image source here

    In doing so you must reduce #Deltax#. If #Delta x->0# you'll get the tangent in P whose inclination will give the inclination at P and thus the derivative at P.

    enter image source here

    Hope it helps!

  • Answer:

    We use quotient rule as described below to differentiate algebraic fractions or any other function written as quotient or fraction of two functions or expressions

    Explanation:

    When we are given a fraction say #f(x)=(3-2x-x^2)/(x^2-1)#. This comprises of two fractions - say one #g(x)=3-2x-x^2# in numerator and the other #h(x)=x^2-1#, in the denominator. Here we use quotient rule as described below.

    Quotient rule states if #f(x)=(g(x))/(h(x))#

    then #(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2#

    Here #g(x)=3-2x-x^2# and hence #(dg)/(dx)=-2-2x# and as #h(x)=x^2-1#, we have #(dh)/(dx)=2x# and hence

    #(df)/(dx)=((-2-2x)xx(x^2-1)-2x xx(3-2x-x^2))/(x^2-1)^2#

    = #(-2x^3-2x^2+2x+2-6x+4x^2+2x^3)/(x^2-1)^2#

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

    or #(2(x-1)^2)/(x^2-1)^2#

    = #2/(x+1)^2#

    Observe that #(3-2x-x^2)/(x^2-1)=((1-x)(3+x))/((x+1)(x-1))=(-3-x)/(x+1)# and using quotient rule

    #(df)/(dx)=(-(x+1)-(-3-x))/(x+1)^2=2/(x+1)^2#

Questions