Differentiation Questions?

Question

Find dy/dx
given y = e^(-2x) In(4x)

please show working out thanks

Answer 1

to do this, use the chain rule:

y' = -2e^(-2x) * ln(4x) + [e^(-2x)] * 1/(4x)
y' = e^(-2x) * [1/(4x) -2]


That should be it...

Answer 2

y = e^(-2x) In(4x)
use partial differential method --> u'v +uv'
dy/dx = -2 . e^(-2x) . ln (4x) + e^(-2x) . 4/x
= 2e^(-2x) (2/x-ln (4x))

Answer 3

d(e^(-2x) ln (4x))/dx

Using the product rule:

d(e^(-2x))/dx ln (4x) + e^(-2x) d(ln (4x))

Using the chain rule:

e^(-2x) d(-2x)/dx ln (4x) + e^(-2x)/(4x) d(4x)/dx

Finishing up:

e^(-2x) (-2) ln (4x) + e^(-2x)/(4x) (4)
-2e^(-2x) ln (4x) + e^(-2x)/x

And we are done.