Drawing derivative image in MATLAB How to draw the derivative image of y = sin (x) with plot x=0:10

x=0:0.1:10;
y=sin(x);
z=diff(y);
plot(x(1:end-1),z)

Three A and seven B are true fractions, and three a + seven B are about 1.38, so what is a of B equal to?

2/5

In the parallelogram as shown in the figure, the area of a is 48 square centimeters, the area of B is 15 percent of the parallelogram, and the area of C is 15 percent______ Square centimeter

Let the area of parallelogram be x, the area of C: x-48-15x, = 45x-48, & nbsp; & nbsp; & nbsp; 48 + 15x = 45x-48, 48 + 15x-15x = 45x-48 − 15x, & nbsp; & nbsp; & nbsp; & nbsp; 48 + 48 = 35x-48 + 48, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 35x = 96, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 35x △ 35 = 96 △ 35, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 160, 45 × 160-48, = 128-48, = 80 (square centimeter), a: the area of C is 80 square centimeter, so the answer is: 80

Given that points a (m, 2) and B (2, n) are all on the image with inverse scale function y = x / M + 3, if the line y = mx-n intersects with X axis at C, the coordinates of C about y axis symmetry C 'can be obtained

Substituting the points a (m, 2), B (2, n) into the inverse scale function y = x / M + 3 respectively, listing the equations, finding out the values of M and N, and then drawing and analyzing the solution into the straight line y = mx-n

It is known that the quadratic function y is equal to the minimum value of FX at the point where x is equal to half of T plus two - the square of a quarter of T, (t is not equal to zero) and f (1) = 0. (1) find the expression of y = f (x) (2) if the minimum value of the function y is equal to FX on the negative one to half of the closed interval is - 5, find the value of t at this time. What I don't understand is that since the expression has been obtained, what is the meaning of the second problem

Suppose f (x) = ax ^ 2 + BX + C
The A, B, C you solve should be an algebraic expression with T
That's why there's a second question

The shadow is a right triangle, its area is 10 square centimeter, find the area of the blank part

How to answer without a picture?

A point P in the second quadrant of the left and right focus F1F2 of the hyperbola x ^ 2-y ^ 2 / 4 = 1 is on the hyperbola, and the coordinates of point P are obtained
As shown in the figure, a point P in the second quadrant of the left and right focus F 1 F 2 of the hyperbola x ^ 2-y ^ 2 / 4 = 1 is on the hyperbola, and ∠ f1pf2 = π / 3, calculate the coordinates of point P

According to the questions a = 1, B = 4, C = radical 5
From pf2-pf1 = 2
(F1F2) ^ 2 = (Pf1) ^ 2 + (PF2) ^ 2-2pf1pf2cos60 degrees
The solution of the equations is PF2 = 1 + root 17
The second definition of hyperbola is (1 + radical 17) / (a ^ 2 / C - x) = E
The solution is x = - radical 85 / 5, y = + - 4 radical 15 / 5
So p (- radical 85 / 5, + - 4 radical 15 / 5)

On the partial derivative of Higher Mathematics
Let R ^ 2 = x ^ 2 + y ^ 2 ------ 1 x = rcost ------ 2 y = rsint -------- 3
If we calculate according to Formula 1: (partial R, partial x) = x / R, if we calculate according to formula 2: (partial R, partial x) = 1 / cost = R / x, we get opposite results according to two different formulas, and the first calculation is correct. Now what's wrong with the second algorithm?
If we calculate according to formula 2: r = (1 / CoS) x, take x as a variable, t as a constant, (partial R, partial x) = 1 / cost = R / x, and do not take the partial derivative as a division, where is it?

Yes, the calculation of Formula 1 is correct. But why do you have to make the two results the same when you calculate the partial derivative of Formula 1 and formula 2 to x according to the implicit function?
Formula 1 is the function of R and X, y, and formula 2 is the function of R and X, T. the two formulas are not the same function. Why do they seek the partial derivatives of X for R respectively, and the results are the same?

Fill in the words as you like. (urgent! ~ ~) gods help
If you want all the words in the group to be the same, please fill in the unified answer of the above group

Face () courtesy () ground () outside () answer: appearance () struggle () encourage () fast () labor answer: diligence

It is known that: the first-order function y = 2X-4. (1) draw the image of the first-order function y = 2X-4 in the rectangular coordinate system. (2) find the area of the triangle formed by the image of the function y = 2X-4 and the coordinate axis. (3) when x takes what value, Y > 0

(1) The intersection point of the first-order function y = 2X-4 and the coordinate axis is (2,0), (0, - 4), as shown in the figure: (2) from the image, we can see that: the area of the triangle = 2 × 4 △ 2 = 4 (3) from the meaning of the problem, Y > 0 is the image, and then above the X axis: 2X-4 > 0, the solution is: x > 2, that is, when x > 2, Y > 0

Drawing derivative image in MATLAB How to draw the derivative image of y = sin (x) with plot x=0:10