(adsbygoogle = window.adsbygoogle || []).push({}); 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

b)

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

c)

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]

d)

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 .

y=mx+b

m= -3/2

y= 2/ 3x-2

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

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

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

y'=-2/3x^-2

now add m which equals y' to find x value

-3/2 = -2/3x

(-3/2)/(-2/3)=x

x=9/4

now to find y value

y= 2 / 3(9/4) -2

y= 2 / (27/4 - 8/4)

y= 2/ (19/4)

y=8/19

now to find b to know the equation

y=mx+b

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 .

m=7

y=x^3 -3x-4

y'=3x^2 -3

now to substitute m which equals y' to find x

7=3x^2 -3

10/3=x^2

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

y=8/3

now to find the equation by finally find b

y=mx+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.

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# Homework Help: Derivatives,First Principle andEquation of a Tangent to a Curve

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