Free Calculus Tutoring

Introduction:

British mathematician, Isaac Newton and the German man Gottfried Leibnitz, invented the calculus. Calculus has  two classifications: Differentiation calculus, Integration calculus. Calculus is the study of rates of change, area, or volume. In symbol, we require to find f(x) where, d/dx f(x) = g(x). Integration is one of most important study of calculus in mathematics. We require to find f(x) = ∫ g(x) dx.In the trigonometric calculus the derivative of a constant is zero. There is no exact value for the integral.


Examples for free calculus tutoring:

Differential free calculus tutoring:

Free calculus tutoring problem 1:

Find the first and second derivative of f(x) = 8x4 + 7x3 - 6x2 - 5x +3

Solution;

Steps to solve:

• First differentiate f(x) = 8x4 + 7x3 - 6x2 - 5x +3                                  
• we know, d / dx (x n ) = n xn-1
• when we differentiate the first term we get  (8×4) x (4-1)
• The above equation can above simplified as 32 x 3
• Like wise we can differentiate  and simplify all the terms in the equation
• Finally after the first derivative we will get  32 x 3 + 21 x2 -12x -5
• Now we will go for Second derivative
• Second derivative of  f(x) = 32 x 3 + 21 x2 -12x -5
• Differentiate all terms   (32 × 3)x2 + (21 × 2) x - (12 × 1) x0
• End of the second derivative we will get   96x2 + 42 x - 12

First derivative:

f' = df / dx

= (8 × 4) x (4-1) + (7 × 3)x(3-1) - (6 × 2)x(2-1) - (5 × 1)x (1-1)

= 32 x 3 + 21 x2 – 12x -5

Second derivative:

f '' = df ' / dx

= 32 x 3 + 21 x2 – 12x -5

= 96x2 + 42 x - 12

Answer for the given free calculus tutoring problem is:

`d^2/dx^2 ` (8x4 + 7x3 - 6x2 - 5x +3) = 96x2 + 42 x - 12



Free calculus tutoring problem  2:

Find the first and second derivative of sin 2x.

Solution;

Steps to solve:

• First differentiate f(x) = sin 2x                                                       
• (we know, d/dx (sin x) = cos x)
• when we differentiate sin x  = cos x So, we get  sin 2x = cos 2x
• Then we differentiate 2x  So, we get  2
• The differentiation of sin 2x = 2 cos 2x
• Finally after the first derivative we will get  2 cos 2x
• Now we will go for Second derivative
• we know,  d/dx (cos x) = −sin x
• The second derivative 2 cos 2x = 2 ×2 (- sin 2x)
• So, The second derivative of  sin 2x = -4 sin 2x

First derivative:

f' = df / dx

= `d/dx` (sin 2x)

= 2 cos 2x

Second derivative:

f '' = df ' / dx

= `d /dx` (2 cos 2x)

= -4 sin 2x

Answer for the given free calculus tutoring problem is:

`d^2/dx^2 `(sin 2x) = -4 sin 2x

I like to share this calculus problems with you all through my article.

Integral free calculus tutoring:


Free calculus tutoring problem 1:

Integrate  `int ` (-2x2 + 8x3 + 2x4 + x5) dx

Solution;

Steps to solve:

• First seperate each term ` int` (- 2x2 + 8x3 + 2x4 + x5) dx
• So, we get `int` (- 2x2 +8x3 + 2x4 + x5) dx =  `int ` - 2x2 dx + ` int ` 8x3 dx + `int` 2x4 dx + `int`   x5  dx
• Integrate first term ` int ` - 2x2 dx =  -2 (1/3) x3                                                                  
•  [ we know,   `int` xn dx = 1/ n+1 (xn+1)]
• Integrate all term and we get   (- 2/3) x3 + (8/4) x4 + (2/5) x5+(1/6) x6
• Finally after the integration we will get  (- 2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

` int ` (-2x2 + 8x3 + 2x4 + x5) dx =` int` -2x2 dx + `int` 8x3 dx + ` int ` 2x4 dx +` int` x5  dx

= -2 `int ` x2 dx +  8  `int` x3 dx + 2` int`x4 dx + `int` x5  dx

= -2 (1/3) x3 + 8(1/4) x4 + 2(1/5) x5+(1/6) x6

= (-2/3) x3 + (8/4) x4 + (2/5) x5+(1/6) x6

= (-2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

Answer for the given free calculus tutoring problem is:

` int` (-2x2 + 8x3 + 2x4 + x5) dx = (-2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6