When an object moves in a horizontal straight line, the distance (unit: m) after T seconds from the initial point is s (T) = 5t-2t ^ 2, then what is the initial velocity of the object? When does it have a speed of 0?

The derivative of s to t is V, so V (T) = s' (T) = 5-4t
Therefore, the initial speed is V (0) = 5-0 = 5m / s
When V (T) = 0, t = 1.25
That is, after 1.25 seconds, the object speed is 0

Knowledge of derivatives The derivation of derivative formula (logarithm, exponent, sine and cosine) and derivative algorithm of function is required, and a detailed process is required (it is better not to use other derivative formulas in the process)

Elementary function:
1 (c)'=0
2 (x^a)'=ax^(a-1)
3 (e^x)'=e^x
4 (a^x)'=a^x*(ln a)
5 (ln x)'=1/x
6 (log a (x)) '= 1 / (x * ln a) -- logarithm of x based on 2
7 (sin x)'=cos x
8 (cos x)'=-sin x
9 (tan x)=1/((cos x)^2)
10 (cot x)'=-1/((sin x)^2)
The above formula is valid for the independent variable x in the function definition domain
Derivative algorithm:
(f(x)+/-g(x))'=f'(x)+/- g'(x)
(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)
(g(x)/f(x))'=(f(x)'g(x)-g(x)f'(x))/(f(x))^2

Use derivative knowledge to answer, troublesome experts Let the tangent l at the point (x, y) on the curve y = x ^ 2 + 1 be parallel to the straight line y = 2x + 1 (1) Tangent point (x, y) (2) Equation for finding tangent L

Because the equation of the curve is y = x ^ 2 + 1
So y '= 2x
(1) Because the tangent L is parallel to the line y = 2x + 1
So the slope of the tangent is 2
So 2x = 2
The solution is x = 1
So the tangent coordinate is (1.2)
(2) . the slope of the tangent line has been found to be 2, and the tangent point is (1.2)
The equation for easy tangent is y = 2x

Basic knowledge of derivatives

Derivative function is also named Ji number and derivative (concept in differential). It is a mathematical concept abstracted from the problem of speed change and the tangent problem of curve. It is also called change rate. For example, if a car walks 600 km in 10 hours, its average speed is 60 km / h. but in actual driving

What should we pay attention to when using derivatives to solve optimization problems in life?

1、 Pay attention to the general steps of using derivatives to solve optimization problems. A bank is preparing to set up a new time deposit business. It is predicted that the deposit volume is directly proportional to the square of the deposit interest rate, the proportion coefficient is K (k > o), the loan interest rate is 4.8%, and all the deposits absorbed by the bank can be lent out. Try to determine what the deposit interest rate is, The bank can obtain the maximum income. Analysis: bank income - loan income - deposit interest, so we can set the deposit interest rate, express the bank income as a function of interest rate, and use the derivative to find the maximum value of the function. If the deposit interest is x, there should be x (O, 0.048). According to the topic: the deposit volume is k expansion, the interest payable by the bank is k expansion, and the loan income is 0.o48k expansion, Therefore, the bank's income is y-0.048kx2-kx3. Since Xi '= 0.096 um x-3 walk x, make Xi' = O, get x-0.032 or x-0 (rounded)

Derivative application - optimization problem! An L-long iron wire is cut into two sections and bent into two squares. To minimize the area and of the two squares, what are the lengths of the iron wires at both ends?

Let the length be x, l-x,
Then the side length of the two squares is x / 4, (l-x) / 4
Then the area of the two squares is: x ^ 2 / 16, (l-x) ^ 2 / 16
Their area sum is: x ^ 2 / 16 + (l-x) ^ 2 / 16 = (L ^ 2-2lx + 2x ^ 2) / 16
[use derivative to solve the minimum value of: (L ^ 2-2lx + 2x ^ 2) / 16]
Let f (x) = (L ^ 2-2lx + 2x ^ 2) / 16
f'(x)=-2L+4x
Make f '(x)

When an object moves in a horizontal straight line, the distance (unit: m) after T seconds from the initial point is s (T) = 5t-2t ^ 2, then what is the initial velocity of the object? When does it have a speed of 0?