# Some derivative problems

## Homework Statement

Find the derivatives for the following questions: a) y=x^lnx b) y=(1+x)^1/x c) y=((x-1)^2e^2x)/((x^2+1)^2(x^3-10)^3)

## Homework Equations

Derivatives for exponential equations, the natural logarithm and derivative rules

## The Attempt at a Solution

For a) I came up with an answer of (x^lnxlnx)/x, however, this is not the correct answer as it should be 2(lnx)x^lnx-1. I know where the x^lnx-1 came from, I am just lost on how that 2 got there.

b) is much the same as a) for me, as my answer of -((1+x)^1/xln(1+x))/x^2 is not close to the correct answer of (((1+x)^1/x)/x^2)((x/(1+x))-ln(1+x)). I don't know if my problems here are stemming from simplification or if I simply overlooked something when solving.

For c), basically I have no idea how to solve this one :tongue: My first try was to use the quotient rule, but this resulted in a HUGE amount of work, and got me thinking that there might be an easier way to solve it. Is there an easier way to solve this question, or is the quotient rule the only way to go about it?

Any help you guys can give for these questions will be greatly appreicated, thanks in advance.

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
for random functions u , v we have
(u o v)'= (u' o v).v'
(uv)'=u'v+v'u
(u/v)'=(u'v-v'u)/v²
(u^n)'=nu'u^(n-1)
(exp u)'=u'exp u
(ln u)'= u'/u

okay now all you have to do is practice .until you master it.

PS: (u o v) means u(v(x)) for any x.
and (something)' = derivative of something.

I know all of the derivative rules, my problems here are stemming from actually applying them to the questions I have posted.

A common trick when differentiating a function, say f(x), that has a function of x in the integrand is to write f(x) as e^(ln(f(x))). In problem a) for example, this would be e^(ln(x^ln(x))), which by properties of logs is the same as e^(ln(x)*ln(x)) = e^(ln(x)^2). So when you differentiate this you will use the chain rule and will have to differentiate ln(x)^2-- this is where the 2 comes from.