Question

Can someone help me step by step...

Find the derivative

y=x^(sin(x))

Answer 1

xcos(x)+sin(x) Product Rule

Answer 2

Since we have a variable term in the exponent, we're not allowed to apply the power rule.
So we need to do logarithmic differentiation! :D

Answer 3

oh nevermind

Answer 4

^ that's not right...
and I need help step by step

Answer 5

The idea with a problem like is, we want to get the variable OUT of the exponent position. To do that, we'll take advantage of one of the rules of Logarithms.
So step 1 is: Take the natural log of both sides.

Answer 6

do i use.. d/dx of b^(x)

ln(b)b^(x)

Answer 7

No, that works when we have a CONSTANT for the base.
But since our base is a variable, AND our exponent is a variable, we have no choice but to imply logarithmic differentiation :D

Answer 8

ln is your friend when it comes to exponents

Answer 9

oh okay

Answer 10

So, the natural log of both sides?

Answer 11

Yes, the natural log is the friendliest log, as sheng mentioned ^^ So we want to work with that one.

Answer 12

\[\huge \ln y = \ln (x^{\sin x})\]
You should get something like this, k? :)

Answer 13

both sides do you mean
ln(y)=ln(x^(sin(x))

Answer 14

oh okay

Answer 15

Hah! Fasted gun in the west ! :3

Answer 16

Fastest.. blah fail -_-

Answer 17

LOL! ;)

Answer 18

lol zep

Answer 19

ln(y) would that be 1/y?

Answer 20

\[\huge \log(a^b)=b*\log(a)\]
On the right side, we want to apply this rule of logarithms before we differentiate.
See how that will work out? :O

Answer 21

It will effectively get that variable OUT of the exponent! yay!

Answer 22

1/y=sin(x)*ln(x)

Answer 23

Woops that's a little messy! :) You took the derivative of the left side, and not the right at the same time XD lolol
Simmer down girly :3 one step at a time.

Answer 24

\[\large \ln y = \sin x \ln x\]
So we've simplified it to this I think? :D

Answer 25

oh my bad. product rule?

Answer 26

You were close with the derivative on the left.. One small problem, Since we're taking our derivative with respect to X, anytime we differentiate a DIFFERENT variable (namely y), we'll have to apply the chain rule, multiplying the term by dy/dx (or y', depending on which notation you prefer).

Answer 27

Yes product rule on the right ^^

Answer 28

wait, im lost now.

Answer 29

\[\large (\ln y)'=\frac{ 1 }{ y }(\frac{ dy }{ dx })\]
Remember what happens when you take the derivative of a y term? :)

Answer 30

yeah, i remember. ^ is that what you meant but in words lol?

Answer 31

Too much words? XD Hah sorry.

Answer 32

Yah :3

Answer 33

hahaha.. it makes more sense in equation form..

Answer 34

pshhh

Answer 35

figure out the right side ok? :D

Answer 36

sin(x)*(1/x)+ln(x)*cos(x)

Answer 37

k cool sauce. We just have a couple minor steps to finish up the problem now.

Answer 38

We want to solve for dy/dx. So let's get rid of the 1/y by multiplying both sides by y.

Answer 39

dy/dx=sin(x)*(1/x)+ln(x)*cos(x)*y

Answer 40

Don't forget that the y is being multiplied by the ENTIRE thing on the right side, not just the last term of it.

Answer 41

[sin(x)*(1/x)+ln(x)*cos(x)]*y

Answer 42

Yay.

Answer 43

Ok the last step. Since our original function was given in terms of only X's on the right. We want to try to do that with our answer as well.

Do you remember what we started with? y= something.
Let's plug that (something) in for the y in our solution here.

Answer 44

got it!

Answer 45

[sin(x)/x)+ln(x)*cos(x)]x^(sin(x))

Answer 46

Yayyy team \c:/

Answer 47

my gosh that one was insane! thank you!

Answer 48

So you can see that, it's NOT as straight forward as the problem initially looks. Takes a little bit of maneuvering to work it out :3

Answer 49

yeah, it sure does.