Y = x ^ ex, find the derivative of Y

y= x^(e^x) (1)
LNY = e ^ x LNX (2) / /: log both sides of (1)
Y '/ y = e ^ x (LNX + 1 / x) (3) / / (2) derivation of X on both sides
Y '= x ^ (e ^ x) e ^ x (LNX + 1 / x) (4) / /: final result

RELATED INFORMATIONS

1. The derivative velocity of X + 1 / ex!
2. Derivative of implicit function of equation y = sin (x + y)
3. The derivative of y = e ^ TaNx TaNx is to the power of E and 2 is to the power of TaNx
4. Given SiNx = 2 / 3, find the value of (cosx SiNx / cosx + sin) + (cosx + sin / cosx SiNx)
5. Let (SiNx) = 2-cos2x, then f (COS π / 6) =?
6. If f (SiNx) = 3-cos2x, then f (cos15 degrees)=
7. Let f (x) = xsinx (x ∈ R). If x ∈ (- π / 2,3 π / 2) and f (x) ≥ (- 1 / 2) x, the value range of X is obtained
8. A student studies the function f (x) = xsinx and draws the following four conclusions: 1) the function f (x) increases monotonically on [− π 2, π 2]; 2) there is a constant M > 0, so that | f (x) | ≤ m | x | holds for all real numbers x; 3) the function f (x) has no minimum value at (0, π), but must have a maximum value; 4) the point (π, 0) is a symmetry center of the function y = f (x) image （　　） A. ③B. ②③C. ②④D. ①②④
9. Higher number, y = the derivative of e ^ x (cosx + xsinx),
10. Find the monotone interval of function f (x) = 2 / × + SiNx Is there any wood to be broken?
11. Why is f (x) = (x-k) ex derivative image tilted less than 90 degrees
12. How to find the derivative of y = (x + 1) (x + 2) (x + 3)
13. What is the derivative of y = x + m
14. The derivative of the function y = 12 (ex + e − x) is () A. 12(ex−e−x)B. 12(ex+e−x)C. ex-e-xD. ex+e-x
15. Y = e ^ x y '= e ^ x can be written as y = f (x) and derivative y = f' (x) Y = x ^ 2, y = 2x written as y = f (x) and derivative y = f '(x)
16. What is the derivative of F (x) = a (X & # 178; - x-1) / e ^ x?
17. What is the derivative of F (x) = a (X & # 178; - x-1) / e ^ 2?
18. What is the derivative of E Λ x?
19. What is the derivative of e ^ (1-x ^ 2)?
20. Y = e ^ x + 1 / e ^ X-1 derivative Y = (e ^ x + 1) / (e ^ x-1) so y '= [(e ^ x-1) * (e ^ x + 1)' - (e ^ x-1) '* (e ^ x + 1)] / (e ^ x-1) ^ 2 [which formula got this step? I can't understand it] = [(e ^ x-1) * e ^ x-e ^ x * (e ^ x + 1)] / (e ^ x-1) ^ 2 = (- 2E ^ x) / (e ^ x-1) ^ 2