Derivative of this problem?

Question

What is the derivative of

y= square root of tan5x

Thanks.

Answer 1

y = (tan 5x) ^(1/2)
Let u = tan 5x
du/dx = 5 sec ² 5x
y = u^(1/2)
dy/du = (1/2) u ^(-1/2)
dy / du = 1/ ( 2u^1/2 )
dy/dx = (dy/du) (du/dx)
dy/dx = [1 / (2u^(1/2) ] (5 sec ² 5x)
dy/dx = [1 / 2 (tan 5x)^(1/2) ] (5 sec ² 5x)

Answer 2

i think you change the sq. root into a exponent which gives you tan5x ^ 1/2.
then use the chain rule..