Primitive function of derivative What is the original function of V (T) 9.8t + 6.5

Primitive function of derivative What is the original function of V (T) 9.8t + 6.5

v(t)=9.8t+6.5
dS(t)/dt=9.8t+6.5
S (T) = (1 / 2) 9.8t ^ 2 + 6.5T + C = 4.9t ^ 2 + 6.5T + C (where C = s (0), is a constant)

If the derivative is 1 / x, find the original function

lnx+C

Derivative of primitive function of 1 / (x-1) What is the known f '(x) = 1 / (x-1) f (x) also f'(x)=(cosx)^2 f（X）=? How to find the original function is best explained

f(x) = ln(x-1) +C
f'(x) = cosx^2 = 1/2 * (1 - cos(2x))
f(x) = 1/2 * x - 1/4 * sin(2x) +C

How to find the derivative of a function at a certain point? For example, the original problem is to do the tangent of x ^ 3 through (0,2)

Do this:
y=x³
y'=3x²
Let the tangent point be (a, a 3)
Then the tangent line is y = 3A (x-a) + a? = 3A? X-2a
Substituting point (0,2) to tangent: 2 = - 2A,
The solution is: a = - 1
So the tangent is y = 3x + 2

Find the derivative of the following function at a specified point 1、f(x)=5x^3-2x^2+x-3,x0=0 2、f(x)=x/sinx,x0=π/2 3、y=x*(8-x)^1/3,x0=0

If f (x) = [x (8-x)] (1 / 3), then f '(x) = 1 / 3) {[x (8-x-x)] (2 / 3)} (8-2x) f' (8-x-3)} (8-2x)} (8-2x) f '(0) = 0, if f (x) = x [(8-x) ^ (1 / 3)] (1 / 3), then f' (x) = (1 / 3) {[x (8-x)] (2 / 3)} (8-2x) f '(8-2x) f' (0) = 0, if f (x) = x [(8-x) ^ (1 / 3)] f '(x) = 8-8-8-8-8-8-8-x) ^ (1 / 3)] f' (8-x) x) ^ (1 / 3) - (x / 3) [(8-x) ^ (- 2 / 3)

Understanding of the derivative method of dividing point in piecewise function The derivative function expression of each partition has been obtained. However, one of the methods for finding the derivative of the boundary point in the review book is to find the limit of the derivative function between the boundary point and each partition, so that the left and right derivatives are equal, so as to obtain the derivative at this point. How to understand this method is to say that the one-sided derivative at both ends of an open interval is obtained by seeking the limit of the derivative function under the condition that the function is continuous, But according to the definition of derivative, derivative itself is the limit. To get the unilateral derivative from the limit of derivative function, I don't understand this point. I'm looking for the advice of experts. Thank you! Thank you downstairs. I know that according to the definition of derivative, the left and right limits exist and equal derivatives exist. I mean, only one-sided derivative method is used. In the book, there is a definition of unilateral derivative. What's more, as I said above, first find the derivative expression f '(x) in the open interval. For example, if the function f (x) calculates their derivative f' (a +) at the end point a + or B - of the interval (a, b), we use this method The derivative function expression f '(x) in the interval is obtained by taking the limit when x tends to a + or B - (the premise is that if the limit exists, it means that it is not differentiable at the side), that is, LIM (x - > A +) f' (x) to get the unilateral derivative. How to understand this method, why is the limit at the end point of the derivative function f '(x) obtained is the side derivative? What is the theoretical support point

First of all, the derivative of a function is also a function. There is nothing strange about finding the limit of derivative function. I believe you have learned Lagrange formula when you review the whole book. This formula establishes the relationship between function variation and derivative function, which is a bridge to study function by derivative
If the function f (x) is continuous on [x0, x0 + H] (H > 0) and differentiable in (x0, x0 + H), the Lagrange formula can be written as follows:
[f(x0+△x)-f(x0)]/△x=f'(x0+θ△x) (0

I didn't understand the derivative from the beginning The problem is very simple, to process Oh ~ Find the derivatives of the following functions: (1)y=x^5 (2)y=x^12 (3)y=x^(-3) (4)y=x(0.3) (5)y=x(108) (6)y=cosx What's more, I'd like to ask if f '(x) = any power of X, f' (x) is equal to 2x, or is it just for x ^ 2? By the way, the fifth is x ^ (108) But I don't understand~ But we haven't learned the formula yet

(1) Y '= 5 * x ^ 4 (2) y' = 12 * x ^ 11 (3) y '= - 3 * x ^ (- 4) (4) y' = 0.3 * x ^ (- 0.7) (5) y '= 108 * x ^ 107 (6) y' = - SiNx, it is OK to memorize the derivative formula. The formula of power function is (x ^ n) '= n * x ^ (n-1) the derivative formula of several common functions: ① C' = 0 (C is constant); ② (x ^ n) '=

What are the derivatives of the following functions 1.y=cos （2-4x） 2. Y = Tan (the third power of X is divided by 2) 3. The cubic power of y = (in 2x) 4. Y = square of root x + 1

1 -4sin(2-4x)
2 3x / 2 (COS (x divided by 2 to the third power)) ^ 2
3 3(ln(2x))^2/x
4 x / square of root x + 1
The process is too troublesome. If you have any questions, please ask me again

Find the derivatives of the following functions 1. Y = SiNx + cosx divided into SiNx cosx, that is, y = denominator is SiNx + cosx, molecule is SiNx cosx 2. The nth power cosnx of y = SiNx 3. Sin0 degrees sin30 degrees sin45 degrees sin60 degrees sin90 degrees sin180 degrees sin360 degrees What are the cos0 degrees, cos30 degrees, 45 degrees, 60 degrees, 90 degrees, 180 degrees and 360 degrees? Maybe I didn't express it clearly

One
y=(sinx-cosx)/(sinx+cosx)
=(tanx-1)/(1+tanx)
=Tan (x-45 degrees)
Y '= sec (x-45 degrees) ^ 2
Two
y'=cosnx*ncosx*sinx^(n-1)-sinx^n*nsinnx
=nsinx^(n-1)(cosnx*cosx-sinx*sinnx)
=nsinx^(n-1)*cos(n-1)x
Three
0 0.5 0.707 0.866 1 0 0
1 0.866 0.707 0.5 0 -1 0

Find the derivative of the following function, （1）y=(3x-5)^10 （2）y=(3x+2)sin5x （3）y=e^2x*cos3x （4）y=2^x*e^x

1.10(3x-5)^9 ×3=30(3x-5)^92.(3x+2)'sin5x+sin5x'(3x+2)=3sin5x+5cos5x(3x+2)3.(e^2x)'cos3x+e^2x(cos3x)'=2e^2x×cos3x+e^2x(-3sin3x)e^2x(2cos3x-3sin3x)4.(2^x)'e^x+2^x(e^x)'=(2^xln2)e^x+2^xe^x=2^xe^x(ln2+1...