find the derivatives of the given functions.
g(t)=sqrt t(1+t)/t^2
Is that √ everything, or
√(t(1+t)) / t^2, or
(√t)(1+t)/t^2 ?
Ah, I see from a prior post, we have
g(t) = √t (1+t)/t^2
you can expand that to be
g(t) = t^(-3/2) + t^(-1/2)
so the derivative is trivial:
g'(t) = -3/2 t^(-5/2) - 1/2 t^(-3/2)
= -(3+t) / (t^2 √t)
or you can use the product rule to get
g'(t) = 1/2√t (1+t)/t^2 + √t (1)/t^2 - 2√t (1+t)/t^3
= (1+t)/2t^2√t + 2t/2t^2√t - (4+4t)/2t^2√t
= (1+t+2t-4-4t)/2t^2√t
= -(3+t)/2t^2√t
find the typo so the two solutions agree!
agree thank you steve
To find the derivative of the given function, we will use the quotient rule along with the chain rule. Here's how you can do it step by step:
Step 1: Identify the function and its components:
g(t) = sqrt(t(1+t)) / t^2
Step 2: Apply the quotient rule, which states that for a function f(t) = u(t)/v(t), its derivative is given by:
f'(t) = [v(t) * u'(t) - u(t) * v'(t)] / [v(t)]^2
Step 3: Find the derivatives of the numerator and the denominator separately.
Let u(t) = sqrt(t(1+t)) and v(t) = t^2
u'(t) = (1/2) * (t(1+t))^(-1/2) * (1 + 2t) (using the chain rule)
= (1 + 2t) / [2 * sqrt(t(1+t))]
v'(t) = 2t
Step 4: Substitute the values into the quotient rule formula:
f'(t) = [v(t) * u'(t) - u(t) * v'(t)] / [v(t)]^2
= [(t^2) * (1 + 2t) / (2 * sqrt(t(1+t))) - (sqrt(t(1+t))) * (2t)] / [t^2]^2
= [t(1 + 2t) - 2t * sqrt(t(1+t))] / [2 * t^2 * sqrt(t(1+t))]
Simplifying the equation further might be possible, depending on the desired form of the answer.
So, the derivative of the function g(t) = sqrt(t(1+t)) / t^2 is:
g'(t) = [t(1 + 2t) - 2t * sqrt(t(1+t))] / [2 * t^2 * sqrt(t(1+t))]