Question

if y^x^y=sinx then find dy/dx​

Answer 1

Answer:

Step-by-step explanation:

y^x^y=sin x

taking log on both sides

ln y^x^y= ln sin x

x^y ln y=ln sin x

differentiating with respect to x

x^y*1/y dy/dx + y(x)^y-1 = 1/sin x * cos x

x^y*1/y dy/dx=cot x -y(x)^y-1

dy/dx=cot x -y(x)^y-1 /  x^y*1/y