- #1

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- Thread starter The_Brain
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- #1

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- #2

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Edit:

y = sin(x)^x

ln(y) = x ln(sin(x))

Now differentiate implicitly.

cookiemonster

y = sin(x)^x

ln(y) = x ln(sin(x))

Now differentiate implicitly.

cookiemonster

Last edited:

- #3

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or: (d/dx)sin(x)^x = (d/dx)e^xln(sin(x)) = (ln(sin(x))+xcosx/sinx)sin(x)^x, that's what you got.

- #4

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lny=ln((sinx)^x)

lny=xln(sinx)

then take the derivative and use product and chain rule ;)

dy/dx(1/y)=x(1/sinx)cosx + ln(sinx)

then simplfy

dy/dx(1/y) = xcotx+lnsinx

dy/dx=y[xcotx+lnsinx]

plug in for y

dy/dx= (sinx)^x[xcotx+lnsinx]

message me if you need any explanations.

- #5

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dw soz

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- #6

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Thats a fun one to solve.

how do you integrate (sin(x))^x?

For the second derivative, I got

(((sin(x))^x)*(ln(sinx)+x/tanx))*(ln(sinx)+x*cotx)+((sin(x))^x)*(cotx-cotx/((sin(x))^2)

but got no way of checking

how do you integrate (sin(x))^x?

For the second derivative, I got

(((sin(x))^x)*(ln(sinx)+x/tanx))*(ln(sinx)+x*cotx)+((sin(x))^x)*(cotx-cotx/((sin(x))^2)

but got no way of checking

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- #7

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uh guys, are you even allowed to do ln(sinx), i mean sinx isn't always positive right?

- #8

HallsofIvy

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- #9

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edit: sorry i misread your post

my question is: how do do ln((-1)^n*sinx)?

- #10

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sin(x)^x=(-1)^x*(-sin(x))^x

so sin(x)^x=(-1)^x*exp(x*ln(-sin(x))) but then I'm stuck in the differentiation because of the (-1)^x

- #11

HallsofIvy

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- #12

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When I try to differentiate it using the suggested method which I get stuck because of the (-1)^x for values of x on the intervals of the form [2kPi-Pi,2kPi].

This is frustrating I know, everywhere I look people use the same method, but to me there is something missing , or maybe there is something wrong with my thinking :(

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