Question

Could someone help me solve step by step
y=e^xlnx

Answer 1

\[e^xlnx\]

Answer 2

What do you mean, solve? You've given an equation with two variables. Do you mean, how do you graph it? Or what?

Answer 3

Solving for what? For x? For when x = 0?

When x is 1, y = 0 by definition.

e^1 = e
ln 1 = 0

e * 0 = 0

Answer 4

it says compute the derivatives of the function

Answer 5

Whoa now derivatives is something way different than solving an equation... You'll need to use the product rule here. Do you know the product rule?

Answer 6

product rule...

Answer 7

lol there is an actual question
product rule for this one
use \((fg)'=f'g+g'f\) with \(f(x)=e^x,f'(x)=e^x,g(x)=\ln(x), g'(x)=\frac{1}{x}\)

Answer 8

"first" d "second" + "second" d "first"

Let first = e^x
Let second = ln x

Answer 9

\[\frac{d}{dx}[f(x) \cdot g(x)] \implies f(x) g'(x) + f'(x) g(x)\]

Answer 10

is it (f*g)'=f'*g+f*g

Answer 11

isnt that 2 f*g

Answer 12

Yes if you prefer F's and G's, I prefer non-confusing words

Answer 13

What @lgbasallote wrote is correct in notation form

Answer 14

\[\frac{d}{dx} (uv) = udv + vdu\]

Answer 15

I also prefer this notation \(\frac{d}{dx}\)

Answer 16

Can you find the derivatives of e^x and ln x?

Answer 17

ok

Answer 18

Yes? No? :-)

Answer 19

nope

Answer 20

my professor just started teaching this stuff today

Answer 21

\(\frac{d}{dx}\) ln|x| = 1/x \(\frac{d}{dx}\) x = 1/x dx
Normally you can just treat the dx like it's a unit, like feet, meters, miles, hours, joules, etc.

\(\frac{d}{dx}\) e^x = e^x \(\frac{d}{dx}\) x = e^x dx

So you get e^x * 1/x + ln|x| * e^x

Or simplified and with the "dx" displayed
\[\frac{e^x}{x}+e^xln|x| \ \ \ \ dx\]

Answer 22

You definitely need to learn the product rule, it'll come back many times later in the course, and in Calculus 2, and in Calculus 3.

Answer 23

Also learn this one soon:


Quotient rule:
\[\frac{d}{dt}\frac{high}{low} =

\frac{((low)d(high)-(high)d(low)}

{(low)^2} \]

Answer 24

Again I think words will help you remember when you go to do it in your head. Symbolism can confuse people who are non-symbolic learners (i.e.: visual learners) when they are first introduced to a complex mathematical concept.

Answer 25

what agentx5 said
product rule
take the derivative of one factor, multiply it by the other factor, then do it the other way around
although it is certainly easier to write
\[(fg)'=f'g+g'f\]

Answer 26

woud the derivative of e^x be e^x

Answer 27

yes

Answer 28

and the derivative of ln x =1/x