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Derivatives,First Principle andEquation of a Tangent to a Curve

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data
    ok so this is going to be divided into 2 parts, 1st is related to First Principle to obtain the derivative,the 2nd part is about obtaining equations of the tangent.

    I'd like to apologize for not being able to write the equations and my work neatly as other threads seem to have, new at this.

    2. Relevant equations
    First Principle Definition of the Derivative
    lim h[tex]\rightarrow[/tex]0 f(a+h)-f(a)/h

    The Power Rule
    f(x)=x^n,n is any real number, then f'(x)=nx^n-1

    The Constant Rule
    f(x)=c, c is any real number, then f'(x)=0

    The Constant Multiple Rule
    f(x)=cg(x),c is any real number, then f'(x)=cg'(x)

    The Sum Rule
    F(x)=f(x)+g(x) then F'(x)=f'(x)+g'(x)

    The Difference Rule
    F(x)=f(x)-g(x) then F'(x)=f'(x)-g'(x)

    The Product Rule
    F(x)=f(x)g(x) then F'(x)=f(x)g'(x)+f'(x)g(x)

    The Chain Rule for Polynomials
    F(x)=(f(x))^n , then F'(x)=nf'(x)f(x)^n-1

    3. The attempt at a solution

    Calculate the derivative of the following functions using the first principle definition of the derivative.

    a) (well this was my 2nd attempt at solving it, i know it isn't right, also i think i screwed up expanding it due to exponential rules, i forgot what i learned for grade 12 advanced functions a few months ago....hahaha i suck, btw im a 26 yo guy redoing grade 12 courses to attempt to go to University for Physics)

    f(x)=x^2 - 2x + 1

    f'(x)= lim h->0 ((x+h)^2 -2(x+h)+1)-(x^2 -2x+1)/h x ((x+h)^2 -2(x+h)+1)+(x^2 -2x+1)/ ((x+h)^2 -2(x+h)+1)+(x^2 -2x+1)
    = ((x+h)^2 -2(x+h)+1)^2 -(x^2 -2x+1)^2 / h((x+h)^2 -2(x+h)+1)
    = ((x^2 +2xh+h^2)-2x-2h-1)^2 -x^4 -4x^2 -1 / h((x+h)^2 -2(x+h)+1)
    = x^4 +4x^2 h^2 +h^4 +4x^2 +4h^2 +1 -x^4 +4x^2 -1 / h(x^2 +2xh +h^2 -2x -2h +1)
    = h^4 +4x^2 h^2 +4h^2 / h(x^2 +2xh +h^2 -2x -2h +1)
    = h(h^3 +4x^2h +4h) / h(x^2 +2xh+h^2 -2x-2h+1)
    = h^3+4x^2 h +4h / x^2 +2xh+h^2 -2x-2h+1
    = h+2x-2h / x^2 -2x+1
    = -h-1 / x^2 +1
    = lim h->0 -(0)-1/ x^2 +1
    = -1/x^2+1

    f(x)=x^3 -3x^2
    f'(x)= lim h->0 ((x+h)^3 -3(x+h)^2)-(x^3 -3x^2) / h x ((x+h)^3 -3(x+h)^2) +(x^3 -3x^2) / ((x+h)^3 -3(x+h)^2)+(x^3 -3x^2)
    = ((x+h)^3 -3(x+h)^2)^2 -(x^3 -3x^2)^2 / h((x+h)^3 -3(x+h)^2)+(x^2 -3x^2)
    = ((x^3 +3x^2 h +3xh^2 +h^3) -3(x^2 +2xh+h^2))^2 -(x^5 -9x^4) / h((x^3 +3x^2 h +3xh^2 +h^3) -3(x2+2xh+h^2))
    =(x^3 +3x^2 h +3xh^2 +h^3 -3x^2 -6xh-3h^2)^2 -x^5 +9x^4 / h((x^3 +3x^2 h +3xh^2 +h^3) -3x^2 -6xh -3h^2)
    =x^5 +9x^4 h^2 +9x^2 h^4 +h^5 -9x^4 -36x^2 h^2 -9h^4 -x^5 +9x^4 / h(x^3 +3x^2 h +3xh^2 + h^3 -3x^2 -6xh-3h^2)
    = h(9x^4 h^2 +9x^2 h^3 +h^4 -36x^2 h -9h^3) / h(x^3 +3x^2 h +3xh^2 +h^3 -3x^2 -6xh -3h^2)
    = 9x^4 h +9x^2 h^3 +h^4 -36x^2 h -9h^3 / x^3 +3x^2 h +3xh^2 +h^3 -3x^2 -6xh-3h^2
    = h+3h +3x^2 +3xh+6x / x^3-3x^2
    = h+3h-1+3xh+6x / x^3
    = Lim h->0 0-3(0)-1+3x(0)+6x / x^3
    = -1+6x / x^3

    f(x)= [tex]\sqrt{2x}[/tex]
    f'(x)= lim h->0 [tex]\sqrt{2x+h}[/tex] - [tex]\sqrt{2x}[/tex] / h x [tex]\sqrt{2x+h}[/tex] + [tex]\sqrt{2x}[/tex] / [tex]\sqrt{2x+h}[/tex] + [tex]\sqrt{2x}[/tex]
    = ( [tex]\sqrt{2x+h}[/tex] )^2 -( [tex]\sqrt{2x}[/tex] )^2 / h( [tex]\sqrt{2x+h}[/tex] + [tex]\sqrt{2x}[/tex] )
    = 2x+h-2x / h( [tex]\sqrt{2x+h}[/tex] + [tex]\sqrt{2x}[/tex] )
    = h / h( [tex]\sqrt{2x+h}[/tex] + [tex]\sqrt{2x}[/tex] )
    = 1/ [tex]\sqrt{2x+h}[/tex] + [tex]\sqrt{2x}[/tex] )
    = lim h->0 1/ [tex]\sqrt{2x+0}[/tex] + [tex]\sqrt{2x}[/tex]
    = 1/ 2[tex]\sqrt{2x}[/tex]

    f(x)= 4/x+1
    f'(x)= lim h->0 (4/x+h+1 - 4/x+1 )/ h
    = (4(x+h+1)-4(x+1) / (x+h+1)(x+1) )/ h
    = 4x+4h+4-4x-4 / h(x+h+1)(x+1)
    = 4/ (x+h+1)(x+1)
    = lim h->0 4 / (x+0+1)(x+1)
    = 4/ (x+1)^2

    Find the points on the curve y=2/3x-2 where the tangent is parallel to the line y= -3/2 x -1 .


    m= -3/2

    y= 2/ 3x-2
    y= 2(3x-2) ^-1
    y'=2/3x^-1 -1

    now add m which equals y' to find x value

    -3/2 = -2/3x

    now to find y value

    y= 2 / 3(9/4) -2
    y= 2 / (27/4 - 8/4)
    y= 2/ (19/4)

    now to find b to know the equation

    8/19 = -3/2 (9/4) +b
    8/19 + 27/8 =b
    b= 577/152

    the equation is y=-3/2x + 577/152 , points are (9/4 , 8/19)

    Find the equation of the tangent to y=x^2 -3x -4 that is parallel to the line y=7x+3 .


    y=x^3 -3x-4
    y'=3x^2 -3
    now to substitute m which equals y' to find x
    7=3x^2 -3
    x= [tex]\sqrt{10/3}[/tex]

    now to find y

    y= ( [tex]\sqrt{10/3}[/tex] )^3 - 3( [tex]\sqrt{10/3}[/tex] )-4
    y=((10/3)^1/2 )^3 - 3(10/3)^1/2 -4
    y=(10/3)^3/2 - 3(10/3)^1/2 -4
    y= -2(10/3)^2/2 -4
    y= -20/3 -4
    y= 80/12 - 48/12
    y= 32/12

    now to find the equation by finally find b


    8/3 = 7( [tex]\sqrt{10/3}[/tex] )+b
    8/3 = [tex]\sqrt{49(10/3)}[/tex] +b
    8/3 = [tex]\sqrt{490/3}[/tex] +b

    so now i stoped here because something just didn't feel right, perhaps i was in the right direction but i just wasn't sure here.
  2. jcsd
  3. Jul 31, 2010 #2
    I will write this in LaTeX. You can see the code by clicking on the image. You wrote

    [tex]f'(x) = \lim_{h \rightarrow 0} \left( \frac{[(x+h)^2-2(x+h)+1] - [x^2-2x+1]}{h} \times \text{something} \right)[/tex]​

    I'm not sure what the something is, since I suspect you might be missing some parentheses. Nonetheless, look at the definition of the derivative again and ask yourself whether it should be there or not. The same applies to (b)-(d) as well.
  4. Aug 1, 2010 #3
    ah!!! you re totally right, i have seen the mistake and fixed a) b) and d) , thank you so much for the help!!!

    the other 2 im just gonna go with what i have, it is completely cool, thank you very much again for the help!!!
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