Question

find the derivative of the y=xlnx/x+lnx

Answer 1

What rules of derivatives do you think we need to use?

Answer 2

I dont know

Answer 3

Just to clarify, it is :
\[\frac{xln(x)}{x} +ln(x) \]
or
\[\frac{xln(x)}{x+ln(x)} \]

Answer 4

second one

Answer 5

Well, since we have a fraction with x in both the numerator and denominator we probably need to start by using the quotient rule.

Answer 6

Do you want to help me in a complete solution to solve it? Please

Answer 7

This is a multi-step derivative and you probably need to work through it, I can help you go through the steps though. I can try and verify answers as well.

Answer 8

ok thanks

Answer 9

Do you remember the quotient rule?

Answer 10

no

Answer 11

If you have a function of x which has x in the numerator and denominator, the numerator and denominator can be thought of as separate functions (with example):
\[f(x)=\frac{g(x)}{h(x)} f(x) = \frac{\sin(x)}{x^2}, g(x) = \sin(x), h(x) = x^2\]

The derivative of f(x) can then be defined as(continuing with example):
\[f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{h(x)^2}, g'(x) = \cos(x), h'(x) = 2x, h(x)^2 = x^4\]
\[f'(x) = \frac{x^2\cos(x) - \sin(x)2x}{x^4} = \frac{x\cos(x) - 2\sin(x)}{x^3}\]

Answer 12

ok but how I can use (ln) in this rule ?

Answer 13

Well, in this case you'll want to separate g(x) and h(x) as well as find their respective derivatives(applying any additional rules as needed, which you are going to need to do)

Answer 14

x+lnx . x.1/x - 1+1/x . xlnx /(x+lnx)^2

this true ?

Answer 15

Is that supposed to be:
\[\frac{(x+\ln(x))(x(\frac{1}{x}-1)+\frac{1}{x}x\ln(x)}{(x+\ln(x)^2}\]
? It's really hard to tell without explicit parentheses.

Answer 16

yes, now its true right؟

Answer 17

That's not quite the answer I got. if:
\[g(x) = x\ln(x) , h(x) = x+\ln(x)\]
What is g'(x)?

Answer 18

(x.1/x) +(lnx) ?

Answer 19

That's correct, and h'(x)?

Answer 20

1+1/x ?

Answer 21

Okay, piecing all the pieces you've said, I get what I have on my paper here, something like:

\[f'(x)=\frac{(x+\ln(x))(1+\ln(x)) - (x\ln(x))(1+\frac{1}{x}) }{(x+\ln(x))^2} \]

Answer 22

true?

Answer 23

No, you can't cancel out the (x+ln(x)) in the numerator with the one in the denominator because of the subtraction in the numerator. It can be simplified down though if you multiply out all your terms in the numerator. It actually simplifies a lot.

Answer 24

ok, what I can do now ?

Answer 25

If you expand the numerator out (x+lnx)(1+lnx)-(xlnx)(1+1/x) becomes:

\[(x+x\ln(x)+\ln(x)+\ln(x)^2)-(x\ln(x)+\ln(x)) \]

Answer 26

how becomes
(x+xln(x)+ln(x)+ln(x)2)−(xln(x)+ln(x))

i don't understand !

Answer 27

(x+lnx)(1+lnx) = x*1 + x*lnx + (lnx)*1 + (lnx)^2 = x + xlnx + lnx + (lnx)^2

(xlnx)(1+1/x) = x(lnx)*1 + x(lnx)/x = xlnx + lnx

Some terms end up cancelling though.

Answer 28

aha ok thanks,Can we simplify more?

Answer 29

Once you cancel the terms in the numerator(due to subtraction) that gets it in the simplest form.

Answer 30

thank you a lot :)