Take the logarithmic derivative

Logarithmic derivative is a method of finding the derivative of a function. The logarithmic operation can reduce the power function, exponential function and power function operation to multiplication operation, and the multiplication operation or division operation can be reduced to addition or subtraction operation, so that the calculation cost of derivative operation is greatly reduced. The logarithmic derivative method is quite widely used.

Chinese name. Logarithmic derivative.

Field. Mathematics.

Function. Find the derivative of the function.

Merit. The derivative operation is computationally intensive to reduce the definition.

For the derivative function, the two sides of the function are first logarithm, and then the derivative is the same, and the derivative is obtained.

This method of derivation is known as the logarithmic derivative [1]. Abbreviated as logarithmic derivative.

Principle. The principle of logarithmic derivative is:

1) Bottom change, ie;

2) Derivative of composite functions, ie.

Applicability. If the function is in the form of product, quotient, radical, power, exponential or exponential function, the logarithmic derivative method is more applicable when finding the derivative, because taking the logarithm can reduce the multiplication or division operation to addition or subtraction operation, and the logarithmic operation can reduce the operation of the root, power function, exponential function and power function to multiplication and division operation.

Derivative examples. 1) Set, ask.

Solve logarithms, find derivatives, so.

2) Set and ask.

Solve logarithms, find derivatives, so.

3) Let the function be determined by the equation and known, and find.

Solve the derivative of both sides of the equation, and get the ,,,.

will be substituted. Note: Since this is a subtraction as a whole, it is useless to take the logarithm first. If it is written as, it is wrong, and logarithms do not have such arithmetic properties.

Application examples: Find the minimum value of the function over the interval and the maximum value of the function on the interval [2].

The sum of solutions is continuous and derivable over the interval, 1) is obtained by taking the logarithm and finding the derivative, so that x (0,1 e) 1 e (1 e,+).

f'(x) Minus 0 Positive.

f(x) Monotonic decrease Minimum: Monotonic increase.

The minimum value of the function on the interval is.

2) Take the logarithm and find the derivation, so, x (0, e) e (e, +

g'(x) Positive 0 Negative.

g(x) monotonically increases maximum monotonically decreases.

The maximum value of the function on the interval is.

This is usually a function of y(x) = u(x) v(x).

The logarithmic derivative method is essentially a chain rule, e.g. y = x x, take the logarithm is log y = xlog x, and then find the derivative of x on both sides at the same time.

The right side must be able to calculate, the left y is a function of x, which is equivalent to log y(x), and the chain rule for the derivative of x is y'y (the independent variable x is omitted here), thus.

y'/y = (xlog x)'

Then you can put the y'Count it out.

y' =y(xlogx)' =x^x(xlogx)'

Since log y is derived to be y'y, so you can always multiply y y at the end by deriving the number and write y';The key question is whether or not taking the logarithm will make your calculations easier.

Memorize the basic derivative formula table (I have summarized the most comprehensive process for you), and let's do a classic example that examines both the definition of the derivative and the derivative formula.

Students, think for a moment and then see the answer.

Students who have read the content of the little brother yesterday will definitely find that this is a derivative problem to find a point, so it must use the definition method. If you can think of this step, you can give praise. But when you really use the definition method to solve the problem, are you disgusted by the first part?

Here the young lady wants to tell you a problem-solving skill. Whenever you see a lot of formulas with root multiplication and division, be sure to remember to take the logarithm and give it a try, and you will find that the world is still very beautiful.

Then we take the logarithm of u.

Is it a light in the eye, then it is very convenient for us to seek guidance.

Let's substitute x=1 to get it.

Let's look at the v part again, let's use the derivative formula directly, either you can't, or it's too troublesome, how troublesome it is, you try it yourself. When x=1, we will send it.

The classic app is inPiecewise functionsAt the junction, if the value of the function and the derivative value are continuous, then the vertical Xun can find the logarithm and then the derivative equal, thus simplifying the operation.

Logarithmic derivative.

It is a way to find the derivative of a function.

method. The logarithmic operation can be a power function and an exponential function.

and power function operations can be downgraded to multiplicative operations, which can be downgraded to addition or quick subtraction operations, so that the amount of derivative operations is greatly reduced.

The logarithmic derivative method is quite widely used.

Derivative Formula:1、c'=0 (c is constant).

2、(xn)'=nx(n-1) (n∈r)。

3、(sinx)'=cosx。

4、(cosx)'=sinx。

5、(ax)'=axina (ln is the natural logarithm.

6、(logax)'=1 (xlNA) (a>0, and a≠1).

7、(tanx)'=1/(cosx)2=(secx)2。

8、(cotx)'=1/(sinx)2=-(cscx)2。

Natural logarithm is to find the logarithm of e, i.e., ln

There are several rules for logarithmic operations.

ln(x*y)=lnx+lny

ln(x/y)=lnx-lny

ln(x^y)=y*lnx

lny=ln

ln(x^2)-ln(x^2-1)+ln(x+2)^(1/3)-ln(x-2)^2^(1/3)

2lnx-ln(x^2-1)+[ln(x+2)]/3-2[ln(x-2)]/3

Natural logarithm: The logarithm based on e is expressed as ln=logex Take the natural logarithm: lnx =2lnx

x (x -1) takes the natural logarithm: ln[x (x -1)]=lnx -ln(x -1)=2lnx-ln(x -1).

Summary. First of all, we need to understand that the logarithmic derivative is taken to facilitate the calculation.

First of all, we need to understand that the logarithmic derivative is taken to facilitate the calculation.

1.Multiply multiple polynomials. 2.

The exponent of the power function has xThe logarithmic method of finding the wild collision derivative is a method of finding the derivative of a function. The logarithmic operation can reduce the power function, exponential dry function and power finger function operation to multiplication operation, and the multiplication operation or division operation can be reduced to addition or subtraction operation, so that the computational cost of derivative operation is greatly reduced.

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Xie Yuan's nuclear envy replied to Tang Chang:

1) lny=xlnx, and the derivation is lost y'y = lnx + 1, so y' =y(lnx+1)=x^x * lnx+1)。

2) bury the void and open lny=sinx ln(cosx), and find the bending year to get y' /y = cosx ln(cosx)+sinx / cosx * sinx)，y' =y[cosxln(cosx)-sinxtanx]

Fang Tong's attack is as follows, please refer to the Zen limbs: Bureau.

Logarithmic derivative, have you learned?

Natural logarithm is to find the logarithm of e, i.e., ln

There are several rules for logarithmic operations.

ln(x*y)=lnx+lny

ln(x/y)=lnx-lny

ln(x^y)=y*lnx

lny=ln

ln(x^2)-ln(x^2-1)+ln(x+2)^(1/3)-ln(x-2)^2^(1/3)

2lnx - ln(x^2-1) +ln(x+2) ]/3- 2[ln(x-2)]/3

Natural logarithm: The logarithm based on e, expressed as ln=logex, and the natural logarithm is taken as lnx =2lnx

x (x -1) takes the natural logarithm: ln[x (x -1)]=lnx -ln(x -1)=2lnx-ln(x -1).

: Knowing y=(x+1)(x+2) (x+3), find y'

Solution: Take the natural logarithm on both sides: lny=ln(x+1)+ln(x+2)-ln(x+3);

Take the derivative of x on both sides to get y'/y=1/(x+1)+1/(x+2)-1/(x+3)

Hence y'=y[1/(x+1)+1/(x+2)-1/(x+3)]=[(x+1)(x+2)/(x+3)][1/(x+1)+1/(x+2)-1/(x+3)]

This makes the calculations much easier.

Here's how:

The phrase "derivative on both sides" omits two words, and should be "derivative of x on both sides".

If: lny is the derivative of y, of course, it is 1 y, but now it is the derivative of x, here since y is a function of x, then apply the derivative of the composite function, first find the derivative of lny for y 1 y, and then multiply by y y for x', i.e., the derivative of lny to x is: y'/y.

When finding the derivative, you should indicate what the independent variable is, otherwise it is easy to make mistakes, where the independent variable is x, and y is a function of x.

As you understand it, the left side is the derivative of y, and the right side is the derivative of x, how can that be correct?

Since y is the dependent variable and a function of x, it cannot be directly equal to cos(x) just like finding the derivative of sin(x), it is equal to sin(x)*x).'=2x*sin(x), where x is considered to be y, which is (siny).'=cosy*(y')=cos(x)*2x, so that (lny) can be understood'=(1/y)*y'Finish.

Natural logarithm: The logarithm of the base of e, expressed as ln=loge

x takes the natural logarithm: lnx = 2lnx

x (x -1) takes the natural logarithm: ln[x (x -1)]=lnx -ln(x -1)=2lnx-ln(x -1).

Parsing: logarithm: log[x].

Natural logarithm: log[x], abbreviated as lnx

Since y is the dependent variable and a function of x, it cannot be directly equal to cos(x) just like finding the derivative of sin(x), it is equal to sin(x)*x).'=2x*sin(x), where x is considered to be y, which is (siny).'=cosy*(y')=cos(x)*2x, so that (lny) can be understood'=(1/y)*y'Finish.

First of all.

Natural pair source number bai

It is to find the logarithm of e.

That is, there are several DU rules for ln logarithmic operations.

So. You should get the idea, dao

lny=ln

ln(x^2)-ln(x^2-1)+ln(x+2)^(1/3)-ln(x-2)^2^(1/3)

2lnxln(x^2-1)

ln(x+2)

2[ln(x-2)]/3