Friday, December 1, 2017

Some considerations in the study of Calculus

Calculus has been invented a few hundred years ago by Sir Isaac Newton and Leibniz Gottfried to study the motion of planets and moons. After the work of these pioneers, several mathematicians have widened the field of Calculus by developing concepts and methods. The applications of Calculus have been extended to the study of phenomena in the physical, biological and social sciences.

Calculus is based on a few simple ideas and these have allowed the development of applications in different fields. The study of Calculus is based on a multi-representational approach to the concepts, methods, and applications represented numerically. analytically and graphically. The interesting element in the study of Calculus is that its core ideas are closed related. For example, the study of limits, derivatives, and integrals form a whole.

Calculus is the study of change and this is best modeled by the study of the behavior of functions. Functions have been studied in Pre-calculus, Different combinations of functions such as addition, multiplication, division and composition have been studied. Other properties of functions have been explored. The study of Calculus is more concerned about the behavior of functions closed to certain points. For example, the study of the different values of a function as the dependent variable comes closer and closer to a certain value leads to the notion of limits. The study of the slope of a tangent line to the graph of a function leads to the notion of derivative. The study of the area between curves leads to the study of integrals.

The study of limits helps in the understanding of the derivatives and integrals. The limit of the slope of a secant line to a curve allows to find the slope of a tangent line to this curve. The slope of the tangent line is the derivative of the function. The limit of the sum of the rectangles of the area between two curves leads to a better approximation of the area between these curves. This leads to the notion of integrals.

Interested in learning more about Calculus visit this site Mathematical Education Center 
.

Thursday, August 3, 2017

Note to the blog readers

Hello Readers,

I have taken a break since the middle of June in publishing posts about Calculus. These posts are additions to the 2 AP Calculus courses that you can visit at Mathematical Education Center. I'll be back soon for great content. You can browse the blog to see previous posts in Calculus, math learning, study skills, etc. If you find the content of this blog useful share it to others. If you want to learn more about Calculus you can subscribe to the courses I mentioned earlier. If you are interested in tutoring in Math, French, ESL and Spanish face-to-face and online visit New Direction Services at www.ndes.biz 

Friday, June 16, 2017

Implicit differentiation

Implicit differentiation involves differentiating implicit functions. An implicit function is an implicit relation between variables. Differentiating an implicit function leads to differentiate the independent variable with respect to the dependent variable. It's basically finding the derivative using the notation dy/dx.

Two methods can be used:

1) You explicit the function

Example 1

Find the derivative of 3xy = 2
Let's explicit the function: y = 2/3x
Let's calculate the derivative: dy/dx = d/dx(2/3x)
                                                          = -2(3x)'/(3x)²
                                                          = -6./9x²
                                                          = -2/3x²
2) If expliciting is not possible, you make transformations in order to find the derivative.

The rules and formulas used to calculate the derivative of different forms of functions apply in the calculations of the derivative of an implicit function.

Since an implicit function is a relationship between the independent and the dependent variable the the application of the derivative rules might seem odd. Let's familiarize ourselves with the derivatives of some expressions where the derivative rules are applied.

Example 2.   Let's y be a function of x find the derivative of y³ with respect to x.
Let's u = y³. we have two functions: u and y. U is a function of y and y is a function of x. U is a composite function. The chain rule has to be applied in order to find the derivative. The formula to apply here is du/dx = du/dy.dy/dx
du/dx = d(y³)/dy.dy/dx
           = 3y²dy/dx.
Example 3 Find the derivative of u = 2x²y
du/dx = d(2x²y)
Let's apply the constant rule
du/dx = 2d(x²y)
 Let's apply the product rule:
du/dx = 2[d/dx(x²)y+x² dy/dx]
          = :2(2xy+x²dy/dx)
du/dx = 4xy+2x²dy/dx 
Example 4 Find the derivative of 3y³+x²y = x-3
Let's differentiate both sides:
d/dx(3y³+x²y) = d/dx(x-3)
3y²dy/dx+2xy+x²dy/dx = 1
3y²dy/dx+x²dy/dx = 1-2xy
(3y²+x²)dy/dx = 1-2xy
dy/dx = 1-2xy/3y²+x²

Practice. Find the derivatives of the implicit functions:
1) x²+y² = 15
2) 3y²-siny = x²
3) x²+2xy-y = 2
Interested in learning more about derivatives and Calculus visit Mathematical Education Center 

           

Saturday, June 10, 2017

Derivative of exponential functions

Derivative of f(x) = b

In the expression above b is a positive real number and is called the base of the exponential function.
The formula to calculate the derivative is d/dx[f(x)] = lnb.bx.

Rule: The derivative of an exponential function is equal to the product of the natural logarithm of the base by the function.

Example 1 calculate the derivative of f(x) = 2x

The given function has the form f(x) = bx. By applying the formula d/dx[f(x)] = lnb. bx  d/dx[f(x)] = ln2.2x

Derivative of f(x) = bu 


Since f is a composite function where u is a function of x the derivative of f is d/dx [f(x)] = d/du(bu).du/dx= lnb.bu .u'


Rule: The derivative of an exponential function with base b is equal to the product of the natural logarithm of the base by the derivative of u.


Example 2. Calculate the derivative of f(x) = 32x

Let’s apply the formula for the derivative of f(x) = bu which is d/dx[f(x)] = lnb.bu.u’
d/dx[f(x)] = ln3.32x(2x)’
                = ln3.32x.2
                = 2ln3.32x
Derivative of f(x) = ex

The derivative f(x) = eis a special case of f(x) = bx where b = e

Let's substitute b in the formula d/dx[f(x)] = lnb.bx
d/dx[f(x)] = lne.ex
Since lne = 1 d/dx[f(x)] = ex

Rule: The derivative of the function f(x) = eis the function eitself. 


Derivative of f(x) = eu 


Since f is a composite function where u is a function of x its derivative is given by the derivative of a composite function.

Then d/dx[f(x)] = d/du(eu).du/dx = eu.u’

Rule: The derivative of the composite exponential function with base e is equal to the product of the composite function by the derivative of the function u.



Example 3. Calculate the derivative of f(x) = e3x2 ( Note this is not e.3x2 but e with the exponent 3x2)
Let’s apply the formula for the derivative of f(x) = eu which is d/dx[f(x)] = eu.u’
d/dx[f(x)] = e3x2.(3x2)’ 
                = e3x2(6x)
                = 6xe3x2


Summary


The derivative of f(x) = bwhere b>0 is d/dx(bx) - lnb.bx

The derivative of the composite function f(x) = bu where u is a function of x is d/dx(bu) = lnb.bu u’
The derivative of f(x) = ex is d/dx(ex) = ex
 The derivative of f(x) = eis d/dx(eu) = eu.u

Practice
Calculate the derivative of the following functions: 3x2
1) f(x) = e6x 
2) f(x) = e3x2-4x+3 ( 3x2-4x+3 is the exponent )
3) f(x) = ex-e-x/ex-e-x

Interested in learning more about Calculus visit this site Mathematical Education Center

Friday, June 9, 2017

Derivative of logarithmic functions

In this post I'll show some techniques to remember the formulas for logarithmic  functions. I'll do some examples and leave some exercises to practice.

Derivative of logarithmic functions 

Derivative of logbx


d/dx (logbx) = 1/xlnb
To remember this formula let's apply the following technique:
1) Multiply the number of which we calculate the logarithm by the natural logarithm of the base. The number here is x and the base is b. Therefore we have xlnb
2) Take the inverse of this product. The inverse of the product is 1/xlnx

Derivative of lnx

d/dx(lnx) = 1/x
The derivative of the logarithm of any number is equal to the inverse of this number.

Derivative of logbu

Since logbu is a composite function its derivative is given by d/dx(logbu) = d/du(logbu).du/dx

d/dx(logbu) = 1/ulnnb.du/dx

Rule: The derivative of the logarithm of a composite function is equal to its derivative with respect to the new variable (u) multiplied by the derivative of the new variable (u) with respect to x.

Derivative of lnu 

Since u is a composite function we have d/dx(lnu) = d/du(lnu).du/dx
                                                                              = i/u.du/dx

Rule: The derivative of the natural logarithm of a composite function u is equal to the inverse of the function multiplied by its derivative with respect to x

Example 1, Calculate the derivative of y = x³log52x

The derivative of y is y" = (x³log52x)'

Let's apply the product rule:
Y' = (x³)'(log52x) + x³(log52x)'
The derivative of x³ is obvious. Let's calculate the derivative  of log52x
Let's write u = 2x we have (log5u)' = d/du(log5u),du/dx
                                                     = i/uln5.u'
                                                    = 1/2x.ln5.(2x)'
                                                   = 1/2x.ln5.2
                                                   = 1/xln5
let's go back to the derivative of y we have:
y' = 3x²log52x + x³.1/xlnx
   = 3x²log52x+x²/lnx

Example 2.  Calculate the derivative of y = ln(2x²-4x+3)

Let's write u = 2x²-4x+3
We have y = lnu
Then dy/dx = d/dx(lnu)
Since lnu is a composite function then dy/dx = d/du(lnu).du/dx
                                                                      = 1/u(4x-4)
Substitute u: dy/dx = (1/2x³-4x+3).(4x-4)
dy/dx = 4x-4/2x²-4x+3
           = 4(x-1)/2x³-4x+3

Practice

Calculate the derivative of the following functions;
1.log₅(2x+5)
2. 5/log(x+4)
3. ln(sinx)

Interested in learning more about Calculus AB visit this site Center for Integral Development 









Saturday, May 27, 2017

Derivative of Trigonometric functions

In this post I will state the formulas for the derivative of trigonometric functions. I will also give some techniques to remember them and solve problems. I'll do some examples and give some exercises for practice. The formulas will not be demonstrated here.

It's not sufficient to know the formulas for the derivative of trigonometric functions to be able to calculate the derivative of functions containing trigonometric expressions. The calculations of these functions involve being able to apply all the other rules that enable to calculate the derivative of a function.

Derivative of the function sine

The derivative of the function sine is equal to the function cosine. If f(x) = sinx f''(x) or d/dx(sinx) = cosx

Derivative of the function cosine

The derivative of the function cosine is equal to the opposite of the function sine. If f(x) = cosx f'(x) or d/dx(cosx) = -sinx

Derivative of the function tangent

The derivative of the function tangent is equal to the square of the secant function. If f(x) = tanx f'(x) or d/dx(tanx) = sec²x

Derivative of the function cotangent

The derivative of the function cotangent is equal to the opposite of the square of the cosecant function. If f(x) = cotx  f''(x) or d/dx(cotx) = -csc²x

Derivative of the function secant

The derivative of the function secant is equal to the product of the function secant by the function tangent. If f(x) = secx f'(x) or d/dx(secx) = secx.tanx

Derivative of the function cosecant

The derivative of the function cosecant is equal to the opposite of the product of the function cosecant by the function cotangent. If f(x) = cosecx f'(x) or d/dx(cosecx) = -cosecx.cotx.

Observations that allow to memorize the formulas

1) All the derivatives of co-functions have the negative sign. For examples, the derivative of cosx = -sinx, the derivative of cotx = -cosec²x, the derivative of cosecx = -cosecx.tanx
2) For the sine and cosine functions the derivative of the first function is equal to the second function The derivative of sinx is cosx. The derivative of the second function is equal to the opposite of the first function. The derivative of cosx is -sinx
3) When thinking about the drivative of the tangent and cotangent functions think about the the square of the function secant and cosecant. The derivative of the tangent goes with the square of the secant Example the derivative of tanx = sec²x. The derivative of cotangent goes with the square of the cosecant preceded by the negative sign, Example the derivative of cotx = -csc²x
4) For the derivative of the functions secant and cosecant think about multiplying the function secant by the function tangent and the cosecant by cotangent. Example the derivative of secx = secx.tanx. The derivative of coscx = -coscx.cotx In the case of the derivative of the cosecant don't forget to place negative placed before the product.  

Example 1

If f(x) = x²cosx+sinx find f'(x)

The derivative of a sum of two functions is equal to the sum of the derivatives of each function.
f'(x) = (x²cosx)'+(sinx)'
 Applying the product rule to calculate the derivative of x²cosx
f'(x) = (x²)'(cosx) + (x²) (cosx)'+ cosx. I apply the formula (uv)' = u'v+uv'
f'(x) =  2xcosx + (x²)(-sinx) + cosx
        =  2xcosx-x² sinx+cosx
        = -x²sinx + 2xcosx + cosx.

Example 2

If f(x) = sin²x find f'(x)
Let's write f(x) as f(x) = (sinx)²
Let's write sinx = u. Then f(x) = u² and f(u) = u²
The function f becomes the function composite f(u)
The derivative of the composite function f(u) is f'(u) = f'(u).u'
Since f(x) and f(u) are both equivalent we have f(x) = f'(u).u'
f'(x) = 2u,u'
       = 2sinx.(sinx)' (Substituting u)
       = 2sinxcosx.


Example 3

Find the derivative of f(x) = sinx-1/sinx+1

Applying the quotient rule f'(x) = (sinx-1)'(sinx+1)-(sinx-1)(sinx+1)'/(sinx+1)²
Calculating the derivatives: f'(x) = cosx(sinx+1)-(sinx-1)cosx/(sinx-1)²
f'(x) = sinxcosx+cosx-sinxcosx+cosx/(sinx-1)²
f'(x) - cosx/(sinx-1)²

Practice

1) What are the techniques to memorize the formulas of the derivative of the following functions
a) sine and cosine
b) tangent and cotangent
c) secant and cosecant

2) Calculate the derivatives of the following functions:
a) f(x) - xsinx+4
b) f(x) = xcox-x²tanx-2
c) f(x) = cos³x
d) f(x) = cosx+sinx/cosx-sinx

Interested in knowing more about derivatives visit this site Center for Integral Development

Tuesday, May 23, 2017

Derivative of a composite function


Derivation of a composite function

Let's consider a function g. The image by g of any element x of its domain is g(x). Let's consider another function f. The image of g(x) by f is f[g(x)] also written as fog(x). The function fog is called the composed function of g and f.

If g is differentiable for any element x and f is differentiable at g(x) fog(x) = f[g(x)] is differentiable at x. The derivative of the function fog is (fog)'(x) = f''[g(x)].g'(x), The demonstration of this formula is not done here.

The derivative of fog is the product of the derivative of fog by the derivative of g.

If u is a function of x and f is a function of u then f(u) is a composite function. By applying the rule above the derivative of f(u) or f'(u) is equal to the derivative of f(u) multiplied by the derivative of u. We write [f(u)]' = f''(u).u'. If we introduce the notation (d) of differentiability we can write d/dx[f(u)] =d/du[f(u)].du/dx.

In practical applications we have a function f to differentiate with respect to x. We then introduce a function u that is a function of x. Now we have the composite function f(u). The diferentiation or derivative of f with respect to x is equal to the derivative of f with respect to u multiplied by the derivative of u with respect to x . This derivative is called the chain rule. There is a chain of operations to do. First we introduce a new function u. Then we calculate the derivative of f as the the composite function f(u) by applying the formula for the derivative of a composite function.

The chain rule holds also the application of the power rule when we work with a complex function with exponents.

The power rule applies by introducing the new function u.

Example 1 

Let's calculate the derivative of the function f(x) = (2x+1)²

To make the computation of the derivative easy we introduce the function u. Then the function f becomes f(x) = u². The derivative of the function f with respect to x is the derivative of the expression u² with respect to x . We write d/dx[f(x)] = d/dx[u²]

By applying the formula for the derivative of a composed function we have d/dx[f(x)] = d/du(u²).du/dx.

By calculating d/du(u²) we obtain d/dx[f(x)] = 2u. u'

Let's substitute u: d/dx[f(x)] = 2 (2x+1)(2x+1)'

By calculating the derivative of 2x+1 we obtain d/dx[f(x)] = 2(2x+1)(2) = 4(2x+1) = 8x+1

Example 2

Calulate the derivative of f(x) = (x²+3x+4)²

Let's write u = x^2+3x+4

d/dx[f(x)] = d/dx(x²+3x+4)²
                =  d/dx(u²)
               =   d/du(u²).du/dx (Applying the formula of the derivative of a composite function)
               =   2u.u'
               =   2(x²+3x+4)(x²+3x+4)' (Substituting u)
               =   2(x²+3x+4)(2x+3)
              =    2(2x³+6x²+8x+3x²+9x+12)
              =    2(2x³+9x²+17x+12)
              =     4x³+18x²+34x+24)

Interested in learning more about the techniques of calculations for derivatives visit this site and subscribe to the Calculus course