You can differentiate this function by using the quotient rule, which tells you that the derivative of afunction that takes the form
#y = f(x)/g(x)#
can be calculated by using
#color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2#, with #g(x) !=0#.
Your original function looks like this
#f(t) = sqrt(t)/(-4t-6)#
You can rewrite this function as
#f(t) = 1/(-2) * sqrt(t)/(2t+3) = -1/2 * sqrt(t)/(2t + 3)#
The derivative of #f(t)# will thus be
#d/(dt)(f) = -1/2 * [([d/(dt)(sqrt(t))] * (2t + 3) - sqrt(t) * d/dx(2t + 3))/(2t+3)^2]#
#f^' = -1/2 * ( 1/(2sqrt(t)) * (2t + 3) - 2sqrt(t))/(2t+3)^2#
#f^' = -1/(2(2t+3)^2) * (1/(color(red)(cancel(color(black)(2)))sqrt(t)) * color(red)(cancel(color(black)(2)))t + 3/(2sqrt(t)) - 2sqrt(t))#
#f^' = -1/(2(2t+3)^2) * (sqrt(t) + 3/(2sqrt(t)) - 2sqrt(t))#
#f^' = -1/(2(2t+3)^2) * (3 - sqrt(t) * 2sqrt(t))/(2sqrt(t))#
#f^' = color(green)(-1/4 * (3 - 2t)/(sqrt(t) * (2t+3)^2))#
