Instruction

1

Write the given logarithmic expression. If the expression uses the logarithm base 10, the recording is shortened and looks: lg b is the decimal logarithm. If the logarithm is in natural base the number e, then write the expression: ln b is the natural logarithm. It is implied that the result of any logarithm is the power to which it is necessary to build a number of bases, to obtain the number b.

2

Solution of logarithm is calculating the relevant level. Before solving a logarithmic expression, is usually required to simplify the. Convert it using known identities, rules, and properties of the logarithm.

3

Addition and subtraction of logarithms of the numbers b and C on the same grounds is replaced by a single logarithm with a product or division of integers b and C, respectively. Apply as needed the most common transformation, the transition formula of the logarithm to another base.

4

Using expressions to simplify a logarithm, consider the existing restrictions. So the base of the logarithm and can only be a positive number not equal to one. The number b must also be greater than zero.

5

But not always, simplifying the expression, we can calculate the logarithm to its numerical form. Sometimes it doesn't make sense, as a degree represent irrational numbers. In this case, Express the degree numbers written in the form of logarithm.

# Advice 2: How to solve double integrals

From the course of mathematical analysis known concept of double integral. Geometrically, the double integral represents the volume of the cylindrical body based on D and bounded by the surface z = f(x, y). Using double integrals to calculate the mass of a thin plate with a given density, the area of a plane figure, the area of the piece of surface, the coordinates of the center of gravity of a homogeneous plate and the other quantities.

Instruction

1

The solution of the double integrals can be reduced to the computation of definite integrals.

If the function f(x, y) is a closed and continuous in a region D bounded by the line y = c and the line x = d, with c < d and a function y = g(x) and y = z(x), with g(x), z(x) is continuous on [c; d] and g(x) ? z(x) on this interval, then calculate the double integral by the formula presented in the figure.

If the function f(x, y) is a closed and continuous in a region D bounded by the line y = c and the line x = d, with c < d and a function y = g(x) and y = z(x), with g(x), z(x) is continuous on [c; d] and g(x) ? z(x) on this interval, then calculate the double integral by the formula presented in the figure.

2

If the function f(x, y) is a closed and continuous in a region D bounded by the line y = c and the line x = d, with c < d and a function y = g(x) and y = z(x), with g(x), z(x) is continuous on [c; d] and g(x) = z(x) on this interval, then calculate the double integral by the formula presented in the figure.

3

If you want to calculate the double integral more complex regions D, the region D is split into parts, each of which represents an area described in paragraph 1 or 2. Calculates the integral for each of these areas, the results obtained are summarized.

# Advice 3: How to solve derivatives

The derivative is one of the most important concepts not only in mathematics but also in many other fields of knowledge. It describes the rate of change of a function at a given point in time. From the point of view of geometry, the derivative at a point is the tangent of the angle of inclination of the tangent to that point. The process of finding is called differentiation, and backward - integration. Knowing a few simple rules, you can calculate derivative of any function, which in turn makes life easier and chemists, and physicists, and even microbiologists.

You will need

- a textbook of algebra for 9 grade.

Instruction

1

The first thing you need to differentiate the functions is to know the basic table of derivatives. It can be found in any mathematical Handbook.

2

In order to solve the problems associated with finding the derivative, you need to learn the basic rules. So, let's say we have two differentiable functions u and v and some constant value C.

Then:

Derivative of constants is always equal to zero: ()' = 0;

The constant is always imposed for the sign of the derivative: (cu)' = cu';

When finding the derivative of sum of two functions, you just need them to differentiate, and the results folded: (u+v)' = u'+v';

When finding the derivative of a product of two functions, you need the derivative of the first function times the second function and add the derivative of the second function multiplied by the first function: (u*v)' = u'*v+v'*u;

In order to find the derivative from a private two functions is necessary, from the product of the derivative of the dividend, multiplied by a function of the divisor, subtract the product of the derivative of the divisor, multiplied by a function of the dividend is divided by divisor function squared. (u/v)' = (u'*v-v'*u)/v^2;

If given a complex function, then multiply the derivative of the inner function and the derivative from the outside. Let y=u(v(x)), then y'(x)=y'(u)*v'(x).

Then:

Derivative of constants is always equal to zero: ()' = 0;

The constant is always imposed for the sign of the derivative: (cu)' = cu';

When finding the derivative of sum of two functions, you just need them to differentiate, and the results folded: (u+v)' = u'+v';

When finding the derivative of a product of two functions, you need the derivative of the first function times the second function and add the derivative of the second function multiplied by the first function: (u*v)' = u'*v+v'*u;

In order to find the derivative from a private two functions is necessary, from the product of the derivative of the dividend, multiplied by a function of the divisor, subtract the product of the derivative of the divisor, multiplied by a function of the dividend is divided by divisor function squared. (u/v)' = (u'*v-v'*u)/v^2;

If given a complex function, then multiply the derivative of the inner function and the derivative from the outside. Let y=u(v(x)), then y'(x)=y'(u)*v'(x).

3

Using the above-obtained knowledge, you can differentiate almost any function. So, let's look at some examples:

y=x^4, y'=4*x^(4-1)=4*x^3;

y=2*x^3*(e^x-x^2+6), y'=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2*x));

Also there are challenges to calculating the derivative at a point. Imagine you are given the function y=e^(x^2+6x+5), you need to find the value of the function at x=1.

1) Find the derivative function: y'=e^(x^2-6x+5)*(2*x +6).

2) Calculate the value of the function at a given point y'(1)=8*e^0=8

y=x^4, y'=4*x^(4-1)=4*x^3;

y=2*x^3*(e^x-x^2+6), y'=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2*x));

Also there are challenges to calculating the derivative at a point. Imagine you are given the function y=e^(x^2+6x+5), you need to find the value of the function at x=1.

1) Find the derivative function: y'=e^(x^2-6x+5)*(2*x +6).

2) Calculate the value of the function at a given point y'(1)=8*e^0=8

Useful advice

Learn the table of elementary derivatives. This will significantly save time.

# Advice 4: How to solve irrational equations

So, what is the difference between the irrational from the rational equation? If the unknown variable to be under the sign of the square root, the equation is considered to be irrational.

Instruction

1

The main method of solving such equations is a method of construction of both parts

**of the equation**in the square. However. naturally, the first thing you need to get rid of the square root sign. Technically, this method is not complicated, but sometimes it can lead to trouble. For example, the equation v(2x-5)=v(4-7). Raising both sides to a square you will get 2x-5=4x-7. Is the equation to solve is not difficult; x=1. But the number 1 will not be a root of this**equation**. Why? The substitute unit in the equation instead of the value X. in the right And the left part will contain expressions that have no meaning, that is negative. This value is not valid for the square root. Therefore, 1 is a third root, and therefore this irrational equation has no roots.2

So, irrational equation is solved using the method of squaring both parts. And after solving the equation, it is necessary to check, to cut off extraneous roots. This substitute was found in the roots of the original equation.

3

Consider one more example.

2+vх-3=0

Of course, this equation can be solved in the same way as the previous one. To transfer a composite

2+vх-3=0

Of course, this equation can be solved in the same way as the previous one. To transfer a composite

**equation**that does not have a square root on the right side and continue to use the method of squaring. obtained to solve rational equation and check the roots. But there is another way, more elegant. Enter a new variable; vх=y. Accordingly, you will get the equation 2y2+y-3=0. That is the usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Then solve the two**equations**vх=1; vх=-3/2. The second equation has no roots from the first we find that x=1. Don't forget to test the roots.# Advice 5: How to solve definite integrals

The solution of the definite integral always comes down to bringing his original expression to the tabular view, which is easy to calculate. The main problem is search of ways of the cast.

## Common solutions

Repeat a textbook of mathematical analysis or higher mathematics that is definite integrals. As is known, the solution of the definite integral is a function whose derivative will give the integrand. This function is called primitive. On this principle the table of basic integrals.

Determine by referring to the integrand which a table of integrals is suitable in this case. It is not always possible to determine immediately. Often, the table view becomes visible only after several transformations to simplify the integrand.

## Method of change of variables

If the expanded function is trigonometric function, the argument which some polynomial, then try to use the method of change of variables. In order to do this, replace the polynomial at the argument of the integrand, for some new variable. The ratio between the old and new variable define the new limits of integration. Differentiation of this expression, locate the new differential in the integral. Thus, you will get a new form of the previous integral is close to or even corresponding to any table.

## A solution of integrals of the second kind

If the integral is the integral of the second kind, which means the vector form of the integrand, you will need to use the rules of moving from data to scalar integrals. One of these rules is the ratio of Gauss. The law lets go of the rotor flux of some vector function to the triple integral of the divergence of this vector field.

## Substitution of limits of integration

After finding the integral you need to substitute the limits of integration. First substitute the upper limit value in the expression for the integral. You will get some number. Then subtract from the resulting number and another number obtained by substituting the lower limit in the integral. If one of the limits of integration is infinity, then by substituting it in the integral must go to the limit and find what seeks expression.

If the integral is two-dimensional or three-dimensional, you have to portray geometrically the limits of integration to understand how to calculate the integral. Indeed, in the case of, say, the three-dimensional integrals the limits of integration may be the whole plane bounding the volume integrated.