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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

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## Thursday, August 3, 2017

## 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

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.

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.

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

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

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

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 = 2x²ydu/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-3Let'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

^{x }

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.b

^{x.}

^{}

^{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) = 2

^{x}

The given function has the form f(x) = b

^{x}. By applying the formula d/dx[f(x)] = lnb. b^{x}d/dx[f(x)] = ln2.2^{x}^{}

**Derivative of f(x) = b**

^{u }

^{}Since f is a composite function where u is a function of x the derivative of f is d/dx [f(x)] = d/du(b

^{u}).du/dx= lnb.b

^{u}.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) = 3

^{2x}.

Let’s apply the formula for the derivative of f(x) = b

^{u}which is d/dx[f(x)] = lnb.b^{u}.u’
d/dx[f(x)] = ln3.3

^{2x}(2^{x})’
= ln3.3

^{2x}.2
= 2ln3.3

^{2x}**Derivative of f(x) = e**

^{x}**The derivative f(x) = e**

^{}^{x }is a special case of f(x) = b

^{x}where b = e

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

^{x}

d/dx[f(x)] = lne.e

Since lne = 1 d/dx[f(x)] = e^{x}^{x}

^{}

**Rule: The derivative of the function f(x) = e**

^{x }is the function e^{x }itself.

**Derivative of f(x) = e**

^{u}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(e

^{u}).du/dx = e

^{u}.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) = e3x

^{2}( Note this is not e.3x

^{2}but e with the exponent 3x

^{2})

Let’s apply the formula for the derivative of f(x) = e

^{u}which is d/dx[f(x)] = e^{u}.u’
d/dx[f(x)] = e3x

^{2}.(3x^{2})’
= e3x

^{2}(6x)
= 6xe3x

^{2}

**Summary**

**The derivative of f(x) = b**

^{x }where b>0 is d/dx(b^{x}) - lnb.b^{x}**The derivative of the composite function f(x) = b**

^{u}where u is a function of x is d/dx(b^{u}) = lnb.**b**

^{u}**u’**

**The derivative of f(x) = e**

^{x}is d/dx(ex) = e^{x}**The derivative of f(x) = e**’

^{u }is d/dx(e^{u}) = e^{u}.u**Practice**

Calculate the derivative of the following functions: 3x

^{2}
1) f(x) = e

^{6x}
2) f(x) = e3x

^{2}-4x+3 ( 3x^{2}-4x+3 is the exponent )
3) f(x) = e

^{x}-e^{-x}/e^{x}-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.

d/dx (log

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

d/dx(lnx) = 1/x

The derivative of the logarithm of any number is equal to the inverse of this number.

Since log

d/dx(log

= i/u.du/dx

The derivative of y is y" = (x³log

Let's apply the product rule:

Y' = (x³)'(log

The derivative of x³ is obvious. Let's calculate the derivative of log

Let's write u = 2x we have (log

= 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' = 3x²log

= 3x²log

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

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

**Derivative of logarithmic functions**

Derivative of logDerivative of log

_{b}xd/dx (log

_{b}x) = 1/xlnbTo 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 log**_{b}uSince log

_{b}u is a composite function its derivative is given by d/dx(log_{b}u) = d/du(log_{b}u).du/dxd/dx(log

_{b}u) = 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 = x³log**

_{5}2xThe derivative of y is y" = (x³log

_{5}2x)'Let's apply the product rule:

Y' = (x³)'(log

_{5}2x) + x³(log_{5}2x)'The derivative of x³ is obvious. Let's calculate the derivative of log

_{5}2xLet's write u = 2x we have (log

_{5}u)' = d/du(log_{5}u),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' = 3x²log

_{5}2x + x³.1/xlnx= 3x²log

_{5}2x+x²/lnxExample 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.

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

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.

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.

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

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

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 = -csc²x4) 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

Labels:
Calculus,
derivative,
math learning,
math tutoring

## 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

Labels:
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## Friday, May 12, 2017

### Derivative computations

The formula lim f(x)-f(x+h)/h when x→h that defines the derivative of a function f implies tedious calculations to calculate the derivative of some types of functions and combinations of functions..

Therefore some formulas have been established to determine the derivatives of a combination of functions and some specific types of functions.

The formulas for the constant function and the power functions are called respectively constant rule and power rule. The formulas for the sum, product and quotient of functions are called respectively addition rule, product rule and quotient rule. The derivative of a composition of 2 functions f and g is called the chain rule. It is an extension of the power rule The trigonometric, logarithmic and exponential functions have their specific formula.

The derivative of an implicit function is called implicit differentiation.

It is essential to memorize the formulas. Otherwise, it would be difficult to calculate the derivatives of these particular functions. Today we are going to limiting ourselves to the learning of the basic formulas: constant, power, sum, product and quotient rule.

The derivative of the function power defined by f(x) = x

The derivative of the product of a function by a constant is equal to the product of the constant by the derivative of the function power.

If f(x) = ax

The derivative of the function f(x) = x can be calculated using the formula for the derivative of the function power. In order to use this formula we have to write f(x) = x as the function power. We write f(x) = x as f(x) = x

By applying the formula for the function power we obtain f’(x) = x

Therefore some formulas have been established to determine the derivatives of a combination of functions and some specific types of functions.

The formulas for the constant function and the power functions are called respectively constant rule and power rule. The formulas for the sum, product and quotient of functions are called respectively addition rule, product rule and quotient rule. The derivative of a composition of 2 functions f and g is called the chain rule. It is an extension of the power rule The trigonometric, logarithmic and exponential functions have their specific formula.

The derivative of an implicit function is called implicit differentiation.

It is essential to memorize the formulas. Otherwise, it would be difficult to calculate the derivatives of these particular functions. Today we are going to limiting ourselves to the learning of the basic formulas: constant, power, sum, product and quotient rule.

**Derivative of a constant****The derivative of the function constant is 0. If f(x) = c the derivative of f(x) is 0. We write: f′(x) = 0.**

**The Power rule**^{n}is equal to n multiplied by x to the power of n-1. The formula is .f’(x) = nx^{n-1}^{}**Derivative of the product of a constant by a function**The derivative of the product of a function by a constant is equal to the product of the constant by the derivative of the function power.

If f(x) = ax

^{n}its derivative is f’(x) = ax^{n-1}**Derivative of the function f(x) = x**The derivative of the function f(x) = x can be calculated using the formula for the derivative of the function power. In order to use this formula we have to write f(x) = x as the function power. We write f(x) = x as f(x) = x

^{1 }By applying the formula for the function power we obtain f’(x) = x

^{1-1 }^{ }f’(x) = x^{0}⇒ f’(x) = 1**Derivative of a sum of functions****If f. g. h,;;; are differentiable for any value of x of their domain the derivative of the sum of these functions is f’+g’+h’+ ....**

**Derivative of the product of 2 functions****If f and g are differentiable for any value x of their domain the derivative of the product f.g is fg’+gf’**

**Derivative of the quotient of 2 functions****If f and g are differentiable for any value of their domain the derivative of the quotient f/g is (f∕g)’ = f’g-gf’∕g**

^{2}^{}These formulas have to be demonstrated and the learners have to do some exercises to apply them. If anyone is interested in learning more subscribe to these courses via this link Free Introductory Calculus Course and Complete Calculus Course^{}

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