Derivative of complex function: h(t)=sqrt(x/(4-x)) h'(t)=?

Answer 1

Fu(k that
I'd just give up

Answer 2

Just apply the chain rule.

h(x) = (x/(4-x))^(1/2)

h'(x) = (1/2) [(x/(4-x))^(-1/2)] [((1)(4-x)-(x)(-1))/((4-x)^2)]

Because of chain rule, the first part in brackets is the derivative of the square root part. The second part between brackets is the derivative of the division inside the square root.

h'(x) = [1/(2(x/4-x)^1/2)] [4/(4-x)^2] or simplifying [1/((x/4-x)^1/2)] [2/(4-x)^2]

This isn't actually so bad. It's just very hard to write fractions and powers with plain text only.

Answer 3

Let g(x) = x/(4-x). Then your function is √(g(x) or [g(x)]^.5. The derivative of the compound function is -.5*[g(x)]^-1.5 * dg(x)/dx; d(g(x)/dx you get by the derivative of products d(uv) = u*dv+v*du. Set u = x, v = (4-x)^-1;

Answer 4

h(t) = (x/(4-x))^(1/2)
h'(t) = (2/(4-x)^2)(x/(4-x))^-(1/2)
h'(t) = (2/(4-x)^2)((4-x)/x)^(1/2)
h'(t) = 2((4-x)/(x(4-x)^4)^(1/2)
h'(t) = 2(1/(x(4-x)^3))^(1/2)
h'(t) = 2/(4-x)sqrt(x(4-x))