differentiate g(t)=3/sqrt(t)
then find the domain of its derivative
I would write it as
f(t) = 3t^(-1/2)
f ' (t) = (-3/2)t^(-3/2)
or
= -3/(2t√t)
just like the original function, the domain is t > 0
To differentiate the function g(t) = 3/sqrt(t), we can use the quotient rule of differentiation.
The quotient rule states that if we have a function h(t) = f(t)/g(t), where f(t) and g(t) are both differentiable functions, then the derivative of h(t) is given by:
h'(t) = [g(t)f'(t) - f(t)g'(t)] / [g(t)]^2.
In our case, f(t) = 3 and g(t) = sqrt(t).
Let's find the derivatives of f(t) and g(t):
f'(t) = 0, since 3 is a constant and its derivative is always zero.
g'(t) = (1/2) * t^(-1/2), using the power rule for differentiation.
Now, substitute these derivatives into the quotient rule formula and simplify:
g'(t) = [(sqrt(t)*0) - (3*(1/2) * t^(-1/2))] / [sqrt(t)]^2
= [- 3/(2*sqrt(t))] / [sqrt(t)]^2
= -3 / (2 * t)
= -3 / 2t.
Therefore, the derivative of g(t) = 3/sqrt(t) is g'(t) = -3/(2t).