Is that √ everything, or
√(t(1+t)) / t^2, or
(√t)(1+t)/t^2 ?
find the derivatives of the given functions.
g(t)=sqrt t(1+t)/t^2
4 answers
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
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