Question

Solve this question...​

Answer 1

Answer:

y = (sinx )cosx

logy = cosx log(sinx)

1/y dy/dx = cosx ×(1/sinx)× (cosx) +log(sinx) (-sinx)

dy/dx = (cos2x(1/sinx)-sinx(logsinx))y

dy/dx =( cos2xcosecx -sinx (log sinx )(sinx)cosx

Answer 2

Answer:

y = (sinx) cosx

logy = cosx log(sinx)

1/y dy/dx = ( cos2×(1/sinx) –sinx(loginx))y

dy/dx = ( cos2×cosecx–sinx(log sinx)

(sinx)cosk