Tonight,after i finished my assignment i was some how bored so,i prepared something useful for anyone who will start calculus.this is my own writing prepared by using the ms word office and all rights are reserved for mr addoow.
Example 1
Find the equation of the tangent line to the curve F(x) =2x^3-4 at the point (1,-2).
Solution
First find the first derivative of the function f(x) =2x^3-4
There fore, we have F`(x) =6x^2
Note=we know from the given points at x=1 and y=-2, lets evaluate the function at x=1, then we have F` (1) =6
There fore, the slope of the curve written as (m) at the point (1,-2) equals m=6.
Recall your earlier algebraic slope point formula.
Y-y1=m(x-x1)
We have y1=-2, x1=1 and m=6, lets do the simple calculation.
y+2=6(x-1)
y+2=6x-6
y=6x-8, that is the equation of the tangent line of our curve at the point (1,-2).
You can check your answer graphically by using a graphic calculator.
Example2 find the derivative of the following function by using the limit definition.
F(x) =3x-1/x^2-1
Solution
Write the fundamental limit formula
F`(x) =Lim f(x+h)-f(x)
h-0 ---------- ,Lim means the limit as h approaches 0.
h h-0
Replace every where you see x in the given function with x+h then we have
f`(x) =lim [3(x+h)-1]/[(x+h)^2-1]-[3x-1/x^2-1]
h-0 ---------------------------------------------
h
also, note we can not replace h with zero because we end up with a zero denominator which is not allowed.
Lets proceed and simplify this expression by distributing the 3 we get
F`(x) =Lim [3x+3h-1]/ [(x+h) ^2-1]-[3x-1/x^2-1]
h-0 ---------------------------------------------
h
We see we have a fraction, so we need to take a common denominator now, we have a common dominator of [(x+h) ^2-1](x^2-1)
F`(x) =Lim (x^2-1) (3x+3h-1)-(3x-1)[(x+h)^2-1]
h-0 -----------------------------------------------
[(x+h) ^2-1](x^2-1)
----------------------------------------------
h
Now, let’s simplify the top by multiplying out, then we can collect the like terms and cancel if there is any cancellation. Okay
F`(x) =Lim [3x^3+3x^2h –x^2-3x-3h+1]-[3x^3+6x^2h+3xh^2-3x-x^2-2xh-h^2+1]
h-0 ------------------------------------------------------------------------------------
[(x+h) ^2-1](x^2-1)
-------------------------------------------------------------------------------
h
Distribute the Minus sign on the top to get rid of the brackets, we get
F`(x) =Lim 3x^3+3x^2h –x^2-3x-3h+1-3x^3-6x^2h-3xh^2+3x+x^2+2xh+h^2-1
h-0 ----------------------------------------------------------------------------------
[(x+h) ^2-1](X^2-1)
-----------------------------------------------------------------------------------
h
Note=by canceling I mean the result will be zero for example 2+(-2)=0,so they cancelled eachother.
now lets see what else we can do about the top part, we see that (3x^3) and (-3x^2) cancel each other,also,(-x^2) cancels (x^2),sidoo kale (-3x) waxay la baxaysaa (3x) haye maxaa kale,well (+3x^2h) iyo (-6x^h) are like term and when we add we end up with (-3x^2h) and last one (+1) waxay la baxaysaa (-1),so we are left with
F`(x) =Lim -3x^2h-3xh^2+2xh-3h+h^2
h-0 -------------------------------
[(x+h) ^2-1](x^2-1)
---------------------------
h
we notice the upper has has a common factor? Can u see it.yeah,it has a common factor of “h”.so lets factor out. We have
f`(x)=Lim h(-3x^2-3xh+2x-3+h)
h-0 -------------------------
[(x+h) ^2-1](x^2-1)
------------------------
h
now, we have a fraction divided by a constant,in this case “h”. recall that the property
a/b
----=a/b X 1/c( X in this case means times.) so,lets use this property and re write the
C expression.
f`(x)=Lim h(-3x^2-3xh+2x-3+h)
h-0 -------------------------- X I/h, (The big X means Times not x).
[(x+h) ^2-1](x^2-1)
We see the two h cancel out, we are left with
f`(x)=Lim (-3x^2-3xh+2x-3+h)
h-0 -------------------------
[(x+h) ^2-1](x^2-1)
Now, we know that h is approaching value closer and closer to zero, but if we replace h with zero, we are not getting a zero denominator any more .lets go and replace everywhere we see h with zero.
f`(x)=Lim (-3x^2-3x(0)+2x-3+0)
h-0 ----------------------
[(x+0) ^2-1](x^2-1)
Simplify again, any thing multiplied by zero is zero, for example,(13 times zero)=0,so in this case -3x times equals zero
We are left with
f`(x)=Lim (-3x^2+2x-3)
h-0 ----------------- = (-3x^2+2x-3)
(X^2-)(X^2-1) ------------------
(x^2-1)^2
There fore f `(x) = (-3x^2+2x-3)
------------------------
(x^2- 1)2
You could have used something called the quotient rule which is easy but understanding this concept is also very crucial in the study of calculus especially the concept of instantaneous rate of change.
Let’s check our answer by doing it the easy way using the quotient rule.
I am not going to do any explanation now,but just remember f `(x) is read as f prime of x.
[f(x)/g(x)]`=g(x).f `(x)-f(x).g`(x)
-------------------------
[g(x)]^2
The question was find the derivative of the function f(x)= 3x-1/x^2-1
Solution
f `(x)= (x^2-1)(3)-(3x-1)(2x)
------------------------------
[x^2-1]^2
simplify the top by multiplying then we have
f`(x)=3x^2-3-[6x^2-2x]
-------------------------
[x^2-1]^2
Distributing the minus sign in the top we have
f `(x)=3x^2-3-6x^2+2x
------------------------
[x^2-1]^2
Add Ur like terms and u r left with
f `(x)=(-3x^2+2x-3)
-------------
[x^2-1]^2
Same answer!!!!!!!
Thanks for reading.!
