The second derivative problem, urgently need to explain, how to find the second derivative of these

If the second derivative is greater than 0, it means that the first derivative is monotonically increasing over the defined domain, that is, the slope of the original function is increasing.

And because the first derivative is less than 0, the original function decreases monotonically in the defined domain.

To sum up, the function decreases and the slope increases, and the change of the function should be slower and slower. This is shown in the figure below.

That's right: when the first derivative is less than 0, if the second derivative is greater than 0, the function changes more and more slowly.

Your teacher is talking about a different situation.

When the first derivative is greater than 0, if the second derivative is greater than 0, the function changes faster and faster.

It sums it up.

If the second-order derivative is greater than 0, then the original function.

In the decreasing interval, the decreasing (changing) is getting slower and slower;

In the increasing interval, the increment (change) is getting faster and faster.

PS: You can analogy a parabola with an opening upward.

It's pretty clear.

Have fun! I hope it can help you, if you don't understand, please ask, I wish you progress! o(∩_o

In fact, if you draw a parabolic diagram, you can clearly see that the diagram is not obvious. Then the concave and convex properties of the graph can be directly analyzed: when the first derivative is greater than zero (single increase), the parabolic slope changes faster and faster, then the second derivative is greater than 0, then it is concave; If the slope of the parabola changes more and more slowly, the second derivative is less than 0, then it is convex.

When the first derivative is less than zero (single minus), the parabolic slope changes more and more slowly, and the second derivative is less than zero, which is concave; If the slope of the parabola changes faster and faster, the second derivative is greater than zero, then it is convex.

1).〔e^(-x)sinx〕'

e^(-x)sinx+e^(-x)cosxe^(-x)(cosx-sinx)

Then, the hall is e(-x)sinx''Pretend to be Sun Yan.

e^(-x)(cosx-sinx)+e^(-x)(-sinx-cosx)

2e^(-x)cosx

2).〔f(x^2)〕'

2xf'(x^2)

f(x^2)〕'

2f'+2xf''Kaimu* (2x).

2f'+4x^2*f''

3).〔cos2x)^2〕'

2cos2x*(-sin2x)*2

4sin2xcos2x

2sin4x

cos2x)^2〕''

2cos4x*4

8cos4x

If the derivative of order is not zero, then this point is an extreme point, 1) the second derivative is greater than zero, and this point is a minimum value;

2) the second derivative is less than zero, and the maximum value of this point;

3) The second derivative is equal to zero, and continue to find the third derivative until the nth derivative that is not zero is obtained, the odd derivative is zero, and the even derivative is not zero, then this point is an extreme point.

You know the implicit function derivative! That is, for the form f(x,y,z)=0, and z can be expressed as z=f(x,y) in a certain area, then (I can't find the symbol of the partial derivative, so I replaced it with &, sorry) &z &x=-fx fz, &z &y=-fy fz, then it can be calculated. First, fx=2x, fy=2y, fz=2z-4, then &z &x=x (2-z), then this result and then find the partial derivative of y, here is the formula of the division structure to find the derivative of the unknown, first the upper and lower unconductive, then subtract the lower and upper unconducted, and then divide by the square of the denominator) then just bring in &z &y, the result is xy (2-z) 3

It is impossible to judge the inflection point from the second derivative of 0 alone, nor can it judge the extreme point.

According to the definition, judging the inflection point needs to judge the positive and negative situation of the second derivative around the point, and when the left and right second derivative signs are different, it can be judged that there is an inflection point, or if there is a third derivative at the point, when the first and second order are 0, the third order is not 0, and the existence of an inflection point can also be judged.

The second derivative is 0, and it is not possible to tell whether the point is an extreme point, which may or may not be.

An example can be given to:

1) For example, if there is a function f(x)=x 4, then the derivative is f'(x) = 4x 3 and the second derivative is f''(x)=12x 2, it can be seen that when x=0, the second derivative f''(x)=0, the function has no inflection point in the defined domain, but has a minima.

2) f(x)=e x-x 2 2, then the derivative is f'(x)=e x-x, and the second derivative is f''(x)=e x-1, we know that when x=0, the second derivative f''(x)=0, but x=0 is not an extreme point.

The x outside the parentheses indicates that the derivative of x is taken.

d 2 x dy 2 represents the second derivative of x for y, which is calculated by finding the first derivative of y for x, and then the derivative of y.

Since dx dy = 1 ke x, the result is a function of x, there is no y, and the derivative of y cannot be directly obtained, so the second derivative is used in the trick:

d^2 x / dy^2

d(1/ke^x) / dy

d(1/ke^x) / dx )×dx / dy)

I don't understand you either, what is the original question? Can you send it to see?

Find dx dy first, and then find the derivative of y, i.e., d(dx dy) dy

First derivative: ((x 2-6x) (1 2)).'=(1/2) *x^2-6x)^(1/2) *x^2-6x)'

x^2-6x)^(1/2) *x-3)

Second derivative: ((x 2-6x) (1 2) *x-3))).'=((x^2-6x)^(1/2))' * x-3) +x^2-6x)^(1/2) *x-3)'

-1/2) *x^2-6x)^(3/2) *x^2-6x)' * x-3) +x^2-6x)^(1/2) *1

(x^2-6x)^(3/2) *x-3)^2 + x^2-6x)^(1/2)

It should be possible to simplify it again.

To find the derivative of the composite function, this can clearly distinguish the inner function and the outer function.

z y = x (f2) (note below: f2 is still a function of x and xy).

z/(∂y∂x)=2x(f2)+x²[2x(f21)+y(f22)]

It can be seen that as long as the first-order conductor of the point exists and the left and right concave and convex properties are consistent.

The intuitive meaning of the second derivative on the image is the concave and convex curves.

When x approaches the x3 point from the left, the figure is a convex function, at this time the second derivative < 0, and when x tends to x3 from the right, the figure is a concave function, and the second derivative is 0, and the left and right second derivatives of a point are unequal, and it is obvious that the second derivative does not exist here (x3 point).