Let y = (sin2x) ^ x, find dy / DX It'd better be more specific

Let y = (sin2x) ^ x, find dy / DX It'd better be more specific

y= (sin(2x))^x
lny = xln(sin(2x))
(1/y)y' = 2xcos(2x)/sin(2x) +ln(sin(2x))
y' =[2xcos(2x)/sin(2x) +ln(sin(2x))] (sin(2x))^x

What is the quadratic integral of ∫ (0 - > 1) DX ∫ (x ^ 2 - > x) (x ^ 2 + y ^ 2) ^ (- 1 / 2) dy in polar coordinates? What is its value? Ask for detailed explanation

This problem is mainly to find y = X ² Polar coordinate equation, rsin θ= r ² cos ²θ, After finishing: r = sin θ/ cos ²θ
Then ∫ (0 - > 1) DX ∫ (x ^ 2 - > x) (x ^ 2 + y ^ 2) ^ (- 1 / 2) dy
=∫[0->π/4]d θ ∫[0->sin θ/ cos ²θ] (1/r)*rdr
=∫[0->π/4]d θ ∫[0->sin θ/ cos ²θ] 1dr
=∫[0->π/4] sin θ/ cos ²θ d θ
=-∫[0->π/4] 1/cos ²θ d(cos θ)
=1/cos θ [0->π/4]
=√2-1

Quadratic integral in polar coordinates: ∫ [0,1] DX ∫ [0,1] f (x, y) dy

∫[0,1]dx∫[0,1] f(x,y) dy
=∫∫ f (x, y) DXDY integral region is rectangular: 0 ≤ x ≤ 1,0 ≤ y ≤ 1
Make y = x and divide the rectangle into two parts,
The polar coordinate equation corresponding to x = 1 is RCOs θ= 1, i.e. r = 1 / cos θ
The polar coordinate equation corresponding to y = 1 is rsin θ= 1, i.e. r = 1 / sin θ
Original formula = ∫ f (RCOs) θ, rsin θ) r drd θ
=∫ [0→π/4] d θ ∫[0→1/cos θ] f(rcos θ, rsin θ) r dr+∫ [π/4→π/2] d θ ∫[0→1/sin θ] f(rcos θ, rsin θ) r dr

The double integral ∫ DX ∫ f (x, y) dy is transformed into a quadratic integral in polar coordinate system

This doesn't have to be converted into polar coordinates
Really, the result should be

Derivative of power exponential function Derivative of y = x ^ SiNx (x > 0) I calculated it like this = SiNx * x ^ (sinx-1) * cosx. The derivative of x ^ 5 is equal to 5 * x ^ 4 The book says take logarithms on both sides of him: ln y = ln x ^ SiNx It's different from my calculation. Why is my calculation wrong

Of course not
These two functions are not the same. One is a power exponential function and the other is a power function
Of course, the derivation is different
You're wrong. You use the derivative method of power function to find the derivative of power exponential function
The derivation of power exponential function can change the exponential relationship into Multiplication relationship by taking logarithm
ln y=ln x^sinx =sinx*lnx
Derivation on both sides y '/ y = cosx * LNX + SiNx / X
y'=(x^sinx)[cosxlnx+(sinx)/x]

Find the derivative of the implicit function, the x power of y = the Y power of X

Take the logarithm xlny = ylnx on both sides and take the derivative of X LNY + XY '/ y = Y / x + y'lnx on both sides multiplied by xyxylny + y'x ^ 2 = y ^ 2 + y'xylnx shift y'x ^ 2-y'xylnx = y ^ 2-xylny to extract the common factor Y' (x ^ 2-xylnx) = y ^ 2-xylnyy '= (y ^ 2-xylny) / (x ^ 2-xylnx) express the derivative of Y with the formula containing y