If e^(x+3logx) then prove that dy/dx=x^2(x+3)e^x....can anyone help me out...plz note 3+logx is in adition with x and is the power of 'e' its not e^x+3logx

Answer 1

that is true, if 'ln' is used, not 'log'. Well, in some books, ln and log seem interchangeable -they really mean ln, not log.
If we rewrite the expression using ln, not log,
e^(x + 3lnx)
we can separate this into:
(e^x)[e^(3lnx)]
note that we can further rewrite and simplify the e^(3lnx) into e^(ln x^3) = x^3. Thus, we have
(e^x)(x^3)
and we use chain rule:
d/dx (f(x)g(x)) = f'(x)g(x) + f(x)g'(x)
which means, if you have two factors which are functions of x (in the problem, they are e^x and x^3), we get the derivative of the first factor then multiply to the second factor, plus the derivative of the second factor then multiply to first factor. Note that
d/dx (e^x) = e^x
d/dx (x^3) = 3x^2
Thus, the derivative is
(e^x)(x^3) + (e^x)(3x^2)
(e^x)(x^2)(x + 3)

hope this helps~ :)

Answer 2

The only thing i have as a doubt is how com e^(lnx^3)=x^3 ?!?!?!?:O

Answer 3

By formula man e^logx is x