In this article we are going to discuss about preparation for calculus problem solution concept. In ...

1. Preparation For Calculus Problem Solution
In this article we are going to discuss about preparation for calculus problem solution concept. In
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Phase I example preparation for calculus problem solution problems:-
Preparation for calculus problem solution problem 1:-
Integrate the given expression with value to x: int16x^4 - 10x^5 dx
Solution:-
Given int 16x^4 - 10x^5 dx.
int16x^4- 6x^5 dx = int 16x^4 dx. - int 10x^5 dx.
= 16int x^4dx. - int 10x^5 dx.
= (16x^5/5) - (10x^6/6) + c.
= 16(x^5)/ (5) - (5x^6)/ (3) + c.
int 16x^4 - 6x^5 dx = 16(x^5)/ (5) - (5x^6)/(3)+ c.
Answer:
int 16x^4 - 6x^5 dx = 16(x^5)/ (5) - 5(x^6)/(3) + c.
Preparation for calculus problem solution problem 2:-
Integrate the given exponential function: int (e^ (16) + sin x)/10 dx
Solution:

2. int (e^(16x)+ sin x)/10dx = 1/10 int e^16x dx + 1/10 int sin x dx
= 1/10 e^ (16x)/ (16) + 1/10 int sin x dx
= 1/10 e^ (16x)/ (16) + 1/10 (-cos x) + c
= e^ (16x)/ (160) - cos x/10 + c
Answer:
e^ (16x)/ (160) - cos x/10 + c.
Phase II examples preparation for calculus problem solution problems:-
Preparation for calculus problem solution problem 1:-
Find the derivative of sin 16x cos 10x with respect to x.
Solution:
Let y = sin 16x cos 10x
= 1/2 cos 10x sin 16x
= 1/2 sin (16x + 10x) - sin (16x - 10x)
= 1/2 sin 26x - sin (5x)
= 1/2 sin 26x + (1/2) sin 5x
dy/dx = 1/2 d/dx sin 26x - 1/2 d/dx (sin 5x)

3. = 1/2 26cos 26x + 1/2 cos 5x
=26/2 cos 26x - 1/2 cos 5x
=13 cos 26x - 1/2 cos 5x
Answer:
d /dx (sin 16x cos 10x) = 13 cos 26x -1/2 cos 5x
Preparation for calculus problem solution problem 2:-
Find the Second derivative of y = (13x^3 - 5x^4) i.e., (d^2y)/ (dx^2)
Solution:
Given: y = 14x^2 - 5x^4
dy/dx = d/dx 14x^3 - 5x^4
= (3*14) x ^ (3-1) - 5(4*1) x^ (4-1)
= 42x^2 - 20x^3
d/dx dy/dx = (d^2y)/ (dx^2) = d/dx (- 20x^3 + 42x^2)
= (-20*3) x^ (3-1) + (42*2) x (2-1)
= - 60 x^2 + 84x^1
= -60x^2 + 84x.
European mathematicians, Isaac Newton and Gottfried Leibnitz, invented the calculus. Calculus
methods applied to continuous graphs, curves, or functions. It is the study of rates of change, area,
or volume. It is classified into two types: Differentiatial calculus (determines the rate of change of a
quantity), Integral calculus (finds the quantity where the rate of change is known). Calculus has
many applications in engineering, science, and economics.
Example Problems to study calculus problems
Studying https://www.math.ucdavis.edu/resources/learning/tutors/ differential
http://privatetutor.info/ calculus problems of Trigonometric Functions:
Problem: Differentiate y = sin(2x) + cos2x.
Solution:
To avoid using the chain rule, recall the trigonometry identity sin(2x) = 2sinx cosx, and first rewrite
the problem as
y = sin(2x) + cos2x

4. = 2sinx cosx + cosx cosx
Now apply the product rule twice. Then
y I= (2sinx Dcosx + D2sinxcosx) + (cosx Dcosx + Dcosxcosx)
= 2sinx(-sinx) + (2cosx)cosx + cosx(-sinx) + (-sinx)cosx
= -sin2x + 2cos2x "" 2sinx cosx
Using the formula cos2x "" sin2x = cos(2x) we can write as follows
= 2(cos2x "" sin2x) "" (2sinx cosx)
= 2(cos(2x)) "" (sin(2x))
= 2cos(2x) ""sin(2x)
Studying differential calculus problems of Functions Using the Quotient Rule
x2
Problem: Differentiate y = ----------
3x - 1
Solution:
Let
(3x "" 1) Dx2 "" (x2) D3x "" 1
yI = -----------------------------------------------
(3x "" 1)2
(3x "" 1) (2x) "" (x2)(3)
= ----------------------------
(3x "" 1)2
6x2 "" 2x "" 3x2
= -------------------------
(3x "" 1)2
3x2 "" 2x
= --------------------

5. (3x "" 1)2
x(3x -2)
= ------------------
(3x "" 1)2
Studying integral calculus problems of trigonometry function:
Problem:Integrate "sin3x dx.
Solution:
Let u = 3x
so that
du = 3dx, or (1/3)du = dx
Substitute into the original problem, getting
"sin3x dx = "sin u(1/3)du
= (1/3) "sin u du
(Use Anti derivative rule 2 from the beginning of this section.)
= (1/3)(-cos u) + C
= -(1/3)cos3x + C
Practice Problems to study calculus problems:
Differentiation of Trigonometric functions:
Problem:Differentiate y = 3sinx "" 4cosx.
Answer:3cosx + 4sinx
Trigonometric Integrals:
Problem:Integrate "tan 5x dx.
Answer:(1/5) ln |sec 5x| + C