Differentiate y = (x^2)(e^x)
dy/dx = (x^2) * d/dx(e^x) + (e^x) * d/dx(x^2)
= (x^2) * (e^x) + (e^x) * (2x)
OK, so I can do that bit nicely, but then they re-arrange the whole thing in this fashion:
=(x^2)(e^x) + (2x)(e^x)
=(x)(e^x) (x+2)
can someone explain what they did here to get it into that form?
How do I do this e^x derivative problem?
2009-06-12 7:36 am
回答 (6)
2009-06-12 7:48 am
✔ 最佳答案
r u really serious in askin' this question???
2009-06-12 2:52 pm
they used the formula d(u.v)=udv+vdu for solving y=(x^2)(e^x)
u=x^2,v=e^x now u subtitute this values in formula...
then in final answer they removed the common terms x,e^x to get
(x)(e^x)(x+2) if u multiply means u wil get the same answer
u=x^2,v=e^x now u subtitute this values in formula...
then in final answer they removed the common terms x,e^x to get
(x)(e^x)(x+2) if u multiply means u wil get the same answer
2009-06-12 2:43 pm
it is the simple factoring actually.
for the x^2 * e^x + 2x * e^x
= x * x * e^x + 2 * x * e^x factor out the x and e^x
x + 2 is left
= x * e^x ( x + 2 )
for the x^2 * e^x + 2x * e^x
= x * x * e^x + 2 * x * e^x factor out the x and e^x
x + 2 is left
= x * e^x ( x + 2 )
2009-06-12 2:43 pm
notice that in both (x^2)(e^x) and 2x(e^x) they have a similarity with an x and an e^x so you can factor it out....I guess it looks better??
so that is how you get (x*e^x) (x+2)
so that is how you get (x*e^x) (x+2)
2009-06-12 2:42 pm
dy/dx = ( 2x ) e^x + e^x (x^2)
dy/dx = x e^x [ 2 + x ]-------------xe^x is a common factor
dy/dx = x e^x [ x + 2 ]
dy/dx = x e^x [ 2 + x ]-------------xe^x is a common factor
dy/dx = x e^x [ x + 2 ]
2009-06-12 2:41 pm
Factorisation. Both terms have a common factor of xe^x, so they were taken out the front of brackets. Expand to confirm.
收錄日期: 2021-05-01 00:25:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090611233637AA7Y0TA