Question

Compute the derivatives
1. y = e^x ln x , (e rasied to the xlnx )

Answer 1

\[\LARGE y = e^{x\ln x}\]

first step ln both sides
\[\LARGE \ln y = \ln e^{x\ln x}\]
\[\LARGE \ln y = x\ln x\]
do you see what i did?

Answer 2

yes

Answer 3

next step is implicit differentiation

Answer 4

Ok what does that mean

Answer 5

you take the derivative of the left side...and you take the derivative of the right side

Answer 6

so I would be finding the derivative of e^x and lnx right?

Answer 7

Careful....

Answer 8

careful about what?

Answer 9

Did you got what lgba told you???

Answer 10

\(\LARGE \ln y = x\ln x\)

Till here you got or not??

Answer 11

yes I got it.

Answer 12

Now take the derivative both the sides..

Do you know what is the derivative of logy???

Answer 13

I'm not sure

Answer 14

\[\frac{d}{dy}Logy = \frac{1}{y}\]

getting??

Answer 15

yes

Answer 16

now take the derivative both the sides but implicitly..

Meaning by taking derivative on LHS it will become:

\[\frac{1}{y}.dy\]

Getting??

Now take the derivative on RHS..
Do you know about product rule in derivative??

Answer 17

thats multiplying right?

Answer 18

yes you have to use product rule there...

Answer 19

\[\frac{1}{y}dy = (x.\frac{1}{x} + lnx).dx\]

Getting till here??

Answer 20

yes

Answer 21

Now multiply both the sides by y what do you get?/
Show me..

Answer 22

Multiply by y both the sides and tell me what you get??

Answer 23

Or...
\[y=e^{f(x)} => y'=f'(x)e^{f(x)}\]

Answer 24

what happened last time?
http://openstudy.com/users/hooverst#/updates/4fff7a07e4b09082c070a70c

Answer 25

or this?
http://openstudy.com/users/hooverst#/updates/4fff61a4e4b09082c0709384

Answer 26

O lol dpalnc I had to leave and go pick up my mom from work

Answer 27

That's what I am thinking @myininaya mam.
But lgbasallote started it from other side so I was just explaining his method..

Answer 28

I'm not sure if I'm solving it righ but would it be y=e^xlnx=y'(x)e^xlnx

Answer 29

No its not..

Answer 30

The correct form of this will be:

\[\large y = e^{xlnx} \rightarrow y' = .e^{xlnx}.\frac{d}{dx}(x.lnx)\]

Answer 31

Just apply the product rule to the last term...

Answer 32

is the anserw e^xlnx=y=x^x

Answer 33

Rewrite it in proper form..

Answer 34

I have provided you the solution above so just apply the product rule there in the last term..

Answer 35

ok so its y=x^x=e^xlnx ?

Answer 36

No its not..

Answer 37

See,
tell me first:

\[\large \frac{d}{dx}(x.lnx) = ??\]

Apply product rule here and then show me..

Answer 38

See, here in my 8th reply I have solved it you can see there..
My 8th reply..

Answer 39

is it 2lnx/x* x^lnx-1

Answer 40

How you got 2??

Answer 41

Product rule says that:

\[\large (uv) = u.v' + v.u'\]

Using this you will get:

\[\large \frac{d}{dx}(x.lnx) = x. \frac{1}{x} + lnx \rightarrow 1 + lnx\]

So,

\[\large \frac{d}{dx}(x.lnx) = 1 + lnx\]

Now substitute this value there and then tell me what you got?

Answer 42

You have to multiply one term with it just one term..

Answer 43

\[y=e^xlnx \rightarrow y' =e^xlnx \times 1+lnx \] ?

Answer 44

You are not getting yet..

See I have written as:

\[\large y' = e^{xlnx} (\frac{d}{dx}(x. lnx))\]

Now just substitute the value of \(\frac{d}{dx}(x.lnx)\) here and tell me what you get..