Help PLEASE!?

Question

look at these steps plz:

ln y = x ln (sin x)

differentiate:
1/y dy/dx = ln (sin x) + x (1/sin x) (cos x)
------how do you go from the above line to the below line?-----
=> dy/dx = (sin x)^x (ln (sin x)) + x (cos x) (sin x)^(x-1)
= (sin x)^(x-1) (sin x ln (sin x) + x cos x)

Answer 1

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

cross multiply the y:

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

now, ln y = x ln (sin x) = ln ({sin x}^x)
this means y = {sin x}^x ..... use that substitution

dy/dx = {sin x}^x [ ln (sin x) + x (1/sin x) (cos x) ]

this is now what you see below the line.
Helped you?


d:

Answer 2

starting with implicit different ion use the operator d/dx( ) to both sides of your equation,thusly:

d/dx( ln y) = d/dx( xln (sinx)
to obtain
dy/dx = yxcos(x) + ysin(x)lnsin(x) where we multiplied both sides by y to solve for dy/dx.

now eleminate y using the oriniginal equation.
you get y =sin(x)^x

now everywhere you see y on the righthandside of the dy/dx equation. substitute this deffinition of y and simplify.

Answer 3

If you recall you started out with
y = (sin(x))^x
and took to log of both sides to facilitate differentiation:
ln y = xln(sin(x))
You came up with
1/y dy/dx = ln (sin x) + x (1/sin x) (cos x)
which becomes
1/y dy/dx = ln(sin(x)) + xcot(x)
Multiplying both sides by y,
dy/dx = y[ln(sin(x)) + xcot(x)]
substituting the original value of y,
dy/dx = ((sin(x))^x)[ln(sin(x)) + xcot(x)]
The last line is superfluous, but you can arrive at it by multiplying by sin(x)/sin(x):
dy/dx = (((sin(x))^x)/sin(x)) [sin(x)ln(sin(x)) + xsin(x)cot(x)]
(((sin(x))^x)/sin(x)) = (sin(x))^(x - 1) and
sin(x)cot(x) = cos(x), so
dy/dx = ((sin(x))^(x - 1))[sin(x)ln(sin(x)) + xcos(x)]
edit:
dy/dx = (sin x)^x (ln (sin x)) + x (cos x) (sin x)^(x-1)
is correct, but unnecessary, and confuses the issue in my opinion.

Answer 4

becuz, if
ln y = x ln (sin x)
then this means,
ln y = ln ((sin x)^x)
or,
y = (sin x)^x

did u get that?
the general rule is: ln (a^b) = b (ln a)
Now using that, look at your 3rd and 4th lines...

Answer 5

NO Idea