Differentiate (2e^-x)lnx (1 Viewer)

laviloki · Feb 11, 2015

This question has stumped me for some reason.

Any help? Thanks

Squar3root · Feb 11, 2015

product rule

laviloki · Feb 11, 2015

Squar3root said:
product rule

Yeah I know that much, but whenever I do it, i'm getting something different from the answer in the book.

Squar3root · Feb 11, 2015

did you get

$\frac{2e^-x}{x} - 2e^-x*ln(x)$

Squar3root · Feb 11, 2015

fail attempt at latex

Squar3root · Feb 11, 2015

2e^-x/x - 2e^-xln(x)

EpikHigh · Feb 11, 2015

(2e^-x)lnx

u = 2e^-x v = lnx
u' -2e^-x v' = 1/x

product rule: vu' + uv' = dy/dx

-2e^-x*lnx + 2e^-x*1/x

= 2e^-2x - 2e^-x*lnx

Squar3root · Feb 11, 2015

EpikHigh said:
(2e^-x)lnx

u = 2e^-x v = lnx
u' -2e^-x v' = 1/x

product rule: vu' + uv' = dy/dx

-2e^-x*lnx + 2e^-x*1/x

= 2e^-2x - 2e^-x*lnx

small typo, should be 2e^(-x)/x

Silly Sausage · Feb 11, 2015

Let $u=2^{-x}$ and let $v=lnx$
$u'=-2e^{-x};v'=\frac{1}{x}$
The product rule: $\frac{d(f(u)g(v))}{dx}=u'v+v'u$
Hence:
$\frac{2e^-{x}}{x}-2lnxe^{-x}$

EpikHigh · Feb 11, 2015

Squar3root said:
small typo, should be 2e^(-x)/x

oops yea