How do you find the derivative of y=(sin x)^x?

Answer 1

first take"In" for both side
Iny =In (sinx)^x
Iny= x In(sinx)
dy/dx (Iny)= dy/dx(x Insinx)
1/y dy/dx = (Insinx + x ((1/sinx) cosx))
dy/dx= y [(Insinx + x ((1/sinx) cosx))]
dy/dx = [(sin x)^x][(Insinx + x ((1/sinx) cosx))]
=[(sin x)^x][(Insinx + xcotx]
note that dy/dx(Inx )=1/x *dy/dx(x)=1/x

Answer 2

Take ln
ln y = x*ln(sinx) and taking the derivative
y´/y = lnsinx +x*cos x/sin x
and y´= (sin x) ^x ( ln sin x + x*cotan x)

Answer 3

y = [sin(x)]^x

To solve this, use logarithmic differentiation. Take the natural log (ln) of both sides,

ln(y) = ln ( [sin(x)]^x )

Use the identity log[base b](a^c) = c log[base b](a)

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

Differentiate implicitly, we get

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

Simplify.

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

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

Multiply both sides by y,

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

But y = [sin(x)]^x, so the final answer is

dy/dx = [sin(x)]^x [ ln(sin(x)) + x cot(x) ]

Answer 4

Start by taking ln of both sides. (Not mulyiplying by ln.)
lny= ln((sinx)^x) = x*ln(sinx)
(1/y)*(dy/dx) = ln(sinx) + x*(1/sinx)*cosx
(dy/dx) = y*ln(sinx) + x*y*cosx/sinx
dy/dx = (sinx)^x*ln(sinx) + x*(sinx)^x*cosx/sinx
dy/dx = (sinx)^x*ln(sinx) + x*cosx*(sinx)^(x - 1)

Answer 5

ln y=x ln (sinx)

y'= (sin x)^x (1+ln x)