Question

Find dy/dx
of y = 3x^2 + 4cot x​

Answer 1

2^(x).e^(3x)sin4x.{ (log2)+3+4cot4x}``logy=xlog2+3x+log sin 4x` ` rArr(1)/(y).(dy)/(dx)=(log2)+3+(4cos4x)/(sin4x) ...

Answer 2

Answer:

Explanation:

For y = u+v

dy/dx = d(u+v)/dx

= du/dx + dv/dx

here , 3x^2 is u where as 4 cotx is v

dy/dx = d(3x^2)/dx + d(4cotx)/dx

= (3*2x^2-1) + 4(-cosec^2x)

= 6x - 4cosec^2 x