Find the slope of the tangent to:

y = 4/(x+2sqrtx) at x=4
I know the answers -3/32 I just want working out :)

y = 4/(x+2sqrtx)
dy/dx = -(4*sqrt(x)+4)/(sqrt(x)*(x^2+4*x)+4*x^2)
when x = 4,
dy/dx = -(4*sqrt(4)+4)/(sqrt(4)*(4^2+4*4)+4*4^2)
dy/dx = -3/32

Edit: You don't even need the chain rule. Just the quotient rule. (And power rule and sum rule of course.)

to fine the slope of the tangent line you have to find the derivative first then substitute X with the given one which is , in this case , x = 4
y' = - 4 [ 1 + (1 / SQRT x) ] / [ x + 2SQRT x]^2 , now sub. with x = 4
y'(4) = slope at 4 = - 6 / 64 = -3 / 32
hope that helped