Question

Derivative

Answer 1

\[\large x^{3}y^{2}=\log_{e} =(x+y)+\sin(e^{x}) \]

Answer 2

Derivative^^

Answer 3

It looks like you're having a little trouble with the equation tool, is this what it's suppose to look like? :o\[\huge x^3y^2=\ln(x+y)+\sin(e^x)\]

Answer 4

Exactly what im thinking,thts a mis print of 2 equal signs! i thought there's a way of dealing with 2 equals!

Answer 5

Oh i see, you typed it the way it's presented to you?
So someone else made the typo? :D hah that sucks.

Answer 6

yeah :|

Answer 7

how would we solve log_{e} ?

Answer 8

So to take a derivative (with respect to x I assume) we have to differentiate IMPLICITLY. All that means is, we need to take a derivative while it's a big mess like this. We don't have the option of solving it in terms of y=f(x).

You're not familiar with the derivative of natural log yet? :D Just trying to get a feel for where yer at.

Answer 9

\[\large 3x^{2}y^{2}+x^{3}2y \frac{dy}{dx}=\frac{d(\ln(x+y))}{dx} +\sin(e^{x})+\cos(e^{x})\]
I can solve till here

Answer 10

I know smt like logx=1/x but i think thats base 10 smt

Answer 11

Looks like you got through the product rule with no trouble, that's cool.

Woops! let's fix the sine derivative a sec.\[\large \frac{d}{dx}\sin(e^x)\quad = \quad \cos(e^x)(e^x)'\]

Answer 12

You take the derivative of the outer function (sine), and then by the chain rule, multiply by the derivative of the inside.
Right? :D

Answer 13

Im confused between when to apply chain rule and product rule,both seem same to me at times

Answer 14

okay supposing I understood this,what next?

Answer 15

Yah hmm let's see how to best explain it :3

Product rule is for the product of functions.
Example:\[\large x^2(2+x)^2\]

Chain rule is when you have a function INSIDE of a function.
Example:\[\large (2+x^2)^2\]

Answer 16

oWo Thanks!!

Answer 17

You could TECHNICALLY always apply the chain rule to any function, it's just not necessary. Here's a quick example.\[\large x^2 \quad = \quad (x)^2\]Taking the derivative of this gives us,\[\large 2(x)^1(x)'\quad=\quad2(x)\]Because the derivative of x is just 1 right? Blah maybe that was unnecessary :) heh.

Answer 18

sugoi :D !

Answer 19

clear :o

Answer 20

As far as the log goes, it is a good one to memorize, it comes up alllll the time.\[\huge \frac{d}{dx}\ln x=\frac{1}{x}\]And you had mentioned that you thought maybe it applies to base 10.

It actually only applies to base e (the natural log).
If you're given a log with a different base, you have to use rules of logarithms to change it into a base of e so we can differentiate it!

Answer 21

okay!

Answer 22

so just 1/x+1/y there^^?

Answer 23

Woops, we don't want to think of it as 2 separate logs. :D It's one big wooden log with x+y in it.
\[\large \frac{d}{dx} \ln(\text{peas + carrots})=\frac{1}{\text{peas + carrots}}(\text{peas + carrots})'\]
After you differentiate the log, you have to remember to apply the chain rule, since we have more than just x inside of the log.

Answer 24

See how the entire contents is placed in one denominator? :o

Answer 25

okay!