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Question : 66 of 160
Marks:
+1,
-0
Solution:
Consider the equation,
y=ex(log‌x) Differentiate above w.r.t x ,
=+ex‌ln‌x ....(I)
Now,
= +y Different above w.r.t x ,
x=+x x+(x−1)y= +x+(x−1)yx+(x−1)y= +(x−1)y+x =ex− +(x−1)ex ln‌x+x Further simplify the above,
x+(x−1)y=(x−1)ex{+ln‌x} +x =(x−1)[+ex‌ln‌x]+x Consider equation (1),
x+(x−1)y =(x−1+x) x+(x−1)y =(2x−1)
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