![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Y = 3x ^ 4 * e ^ 10 find the 10th derivative of Y
The highest order term is 4 times, so the 10th derivative of y = 0
What is the derivative of y = 3x-2?
If you're looking for the first derivative of X
y‘=（3x-2）'
=（3x)'-(2)'
=3-0
=3
What is the derivative of y = 3x (x ^ 2 + 2)?
Is this 3x or 3 times,
Derivative of y = 3x + 2
Directly use the derivative of the power function: the derivative of y = 3 is the slope of the tangent
(3x + 2Y) (4x-5y) - 11 (x + y) (X-Y) + 5xy, where x = 7 / 2 and y = - 5 / 2
(3x+2y)(4x-5y)-11(x+y)(x-y)+5xy
=12x ²- 7xy-10y ²- 11x ²＋ 11y ²＋ 5xy
=x ²- 2xy＋y ²
=(x-y) ²
=(7/2+5/2) ²
=6 ²
=36
The second derivative of Y - the general solution of the first derivative of 2Y - 3Y = e ^ (- 3) That's equal to e ^ (- 3x)
Characteristic equation C ^ 2-2c-3 = 0
c1=-1,c2=3
General solution of homogeneous formula y = C1E ^ - 1x + c2e ^ 3x
Let the special solution of nonhomogeneous form AE ^ - 3x
Bring into the original equation 9ae ^ - 3x + 6ae ^ - 3x - 3aE ^ - 3x = e ^ - 3x undetermined coefficient solution to get a = 1 / 12
Therefore, the equation is y = C1E ^ - 1x + c2e ^ 3x + 1 / 12e ^ - 3x
RELATED INFORMATIONS
- 1. The general solution of the second derivative of Y + the first derivative of 2Y - 3Y = 6x + 1 The corresponding homogeneous equation is y '' + 2Y '- 3Y = 0, and the characteristic equation is γ^ 2+2 γ- 3 = 0, the general solution is y = C1E ^ x + c2e ^ (- 3x). Since 0 is not the root of the characteristic equation, let the general solution of the non-homogeneous equation y '' + 2Y '- 3Y = 6x + 1 be y * = ax + B, substitute it into the equation, and get 2a-3ax-3b = 6x + 1. Therefore, a = - 2, B = - 5 / 3, y * = - 2x-5 / 3, then the general solution of the original equation is y = C1E ^ x + c2e ^ (- 3x) + 6 --- 2x 5 / 3 What I don't understand is "since 0 is not the root of the characteristic equation, let the general solution of the non-homogeneous equation y '' + 2Y '- 3Y = 6x + 1 be y * = ax + B, substitute it into the equation, and get 2a-3ax-3b = 6x + 1, so what does it mean by a = - 2, B = - 5 / 3, y * = - 2x-5 / 3" What 0 is not the root of the characteristic equation? What is 0? How did ax + B come from? How did you bring the general solution y * = ax + B into the equation? What should a and B do? Please give me your advice. I can't understand it
- 2. （x+2y)/2+(3x-y)/3=5 ① 0.2(x+2y)+0.8(3x-y)=4 ② Calculate this binary linear equation, and add money when friends answer it in a minute
- 3. Area enclosed by star line x = ACOS ^ 3T, y = asin ^ 3T
- 4. If the equations of X and y, x = 3-2t, Y-5 = 3T, then use the algebraic formula containing x to express y, which is correct
- 5. Derivative relationship between The displacement of a particle, the velocity is the derivative of the displacement, The above words appear in the content derivative of mathematics elective 2-1 taught by senior two But I don't quite understand the relationship between them
- 6. Monotone interval of function y = root sign (square of x-2x-3)
- 7. F (x) is an odd function defined on R, and f (x + 2) = - f (x). When 0 ≤ x ≤ 1, f (x) = x ^ 2 + X ① find the period ② find the expression at - 1 ≤ x ≤ 0 ③ Find f (6.5)
- 8. It is known that the definition field of function f (x) is all real numbers of X ≠ O. for any X1 and X2 in the definition field, f (x1 · x2) = f (x1) + F (x2), and when x > 1, f (x) > 0, f (2) = 1 (1) Verify that f (x) = f (- x) (2) F (x) is an increasing function on (0, + infinity) (3) Solving inequality f (|x| + 1) < 2
- 9. If the function f (x) = log defined in (- 1,0) takes 2A as the base and X + 1 as the true number If f (x) is greater than 0, the value range of a is
- 10. If the domain of the function f (x) = x2 + 2x + A / 1 is r, the value range of the real number a is (to the process)
- 11. Solution of univariate cubic equation I want to ask about the root formula ~ don't copy it from the encyclopedia ~ I can't understand ~ I've read it! I wonder if the cubic equation can only be solved in some special form? What is that root expressed by those corresponding coefficients?
- 12. A problem about the derivative of univariate function The question is still unclear Taking y as an independent variable and X as a dependent variable, the transformation equation is proved {(dy/dx) * [(dy)^3/d(x^3)]} - 3 {[(dy)^2/d(x^2)] ^2} = x One question: The answer is: [(dy)^3/d(x^3)] = - d/dy{(dx/dy)^(-3)*[(dx)^2/d(y^2)]}*(dy/dx) ={3(dx/dy)^(-4)*[(dx)^2/d(y^2)]^2-[(dx/dy)^(-3)]*(dx)^3/d(y^3)}*（dx/dy）^(-1) =3(dx/dy)^(-5)*[(dx)^2/d(y^2)]^2-(dx/dy)^(-4)*(dx)^3/d(y^3) But how do I figure it out =3(dx/dy)^(-5)*[(dx)^2/d(y^2)]^2-(dx/dy)^(-4)*(dx)^3/d(y^3)*[(dx)^2/d(y^2)] In other words, it is multiplied by one more [(DX) ^ 2 / D (y ^ 2)] I do this: let DX / dy = t So (Dy) ^ 3 / D (x ^ 3) =-d/dy[t^(-3)*t']*t^(-1) =[3t^(-4)*t'-t^(-3)*t'']*t'*t^(-1) =[3t^(-5)*t'^2-t^(-4)*t''*t']*t' I've done it several times. Why is there more t 'in the last step Please kindly answer it. Don't worry. The steps are a little troublesome. Take your time Thanks a lot^^~~
- 13. Find the value range of the following function: y = x + √ (4-x) ²）
- 14. Example of absolute limit 1. Take an example to show that | f | continuity does not mean f continuity 2. For example, function f is discontinuous everywhere, but | f | is continuous everywhere
- 15. The absolute value of X-2 divided by X-2. What is the limit when x approaches 2
- 16. Find the limit: x ^ - 1-e of LIM (x-0) (1 + x) divided by X
- 17. f(x)=[(x^2)*∫ x→a f(t)dt]/(x-a),limx→a F(x)= Using lobida's law, ∫ x tends to a f (T) DT, what is the derivative equal to?
- 18. What is the derivative of F (x) = ex / x
- 19. N-th derivative of y = e ^ x
- 20. The equation of motion of an object is s = T ^ 3 + 2T ^ 2-1. When t = 2,