Deriving Issues (1 Viewer)

[Avenger6]

Hi guys, I am struggling with this question here:

I have listed my answer (well as close as I have got to the answer) beneath the question, however it is incorrect according to the book. I used the product rule to first derive the numerator to get the equation, (1+lnx)/e^x. I then used the quotient rule to get the answer I have listed in the picture above, but it seems I have made a mistake somewhere. For the quotient rule I had the following values:

u=1+lnx
u'=1/x
v=e^x
v'=e^x

Any help is greatly appreciated.

 

[lyounamu]

Hi guys, I am struggling with this question here:

I have listed my answer (well as close as I have got to the answer) beneath the question, however it is incorrect according to the book. I used the product rule to first derive the numerator to get the equation, (1+lnx)/e^x. I then used the quotient rule to get the answer I have listed in the picture above, but it seems I have made a mistake somewhere. For the quotient rule I had the following values:

u=1+lnx
u'=1/x
v=e^x
v'=e^x

Any help is greatly appreciated.

Let f(x) = x . lnx
f'(x) = x. 1/x + lnx . 1
= 1 + lnx

Now

y= (x . ln x)/ e^x
= f(x) / e^x
dy/dx = [e^x . f'(x) - f(x) . d/dx (e^x)] / e^2x
= [e^x . (1 + lnx) - xlnx . e^x] / e^2x
= [e^x + e^x lnx - xlnxe^x] / e^2x
= [e^x (1+lnx-xlnx)] / e^2x
= (1+ lnx - xlnx) / e^x

EDIT: You got the derivative wrong. Just find the derivative of the top one first separately (to make it easier to understand) like I did.

And then substitute it later.

 

[Avenger6]

EDIT: I see where I went wrong...in affect I was finding the second derivate of the numerator...that was my mistake. Thanks for the help.

 

[lyounamu]

EDIT: I see where I went wrong...in affect I was finding the second derivate of the numerator...that was my mistake. Thanks for the help.

v' = derivate of v
u' = derivative of u

I used the quotient rule. When you use that, it should look like this:

dy/dx = (v . u' - u . v' )/ v^2

v = e^x and u = xlnx

v' = e^x
u' = 1 + lnx

 

[tommykins]

vu' - uv'
v²

I'd also like to point out that it MUST be strictly vu' - uv'.

uv' - vu' yields the wrong answer.