Question

Differentiate  (cotx*3)*5/3w.r.t. x​

Answer 1

Answer:

dy/dx = -5x²{ cosec (x³) (cot x³)^5/3 }

Step-by-step explanation:

y=(cotx³)^5/3

let x³=t

then dt/dx=3x²------------(1)

and z=cot t

dz/dt=-cosec(t)*cot(t)------(2)

y=z^5/3

dy/dz=5/3z^(5/3-1)

dy/dz=5/3z^2/3--------(3)

Using the chain rule

dy/dx=dy/dz*dz/dt*dt/dx

=( 5/3z^2/3) { -cosec(t)*cot(t)} *3x²

Putting the values of z and t

dy/dx =[5/3{ cot x³)^2/3 } { -cosec(x³) cot(x³) }*3x²

dy/dx = -5x²{ cosec (x³) (cot x³)^5/3 }