Derivative of f(t)=t^(3/2) log2 sqrt t+1?

1 Answer
Jan 24, 2018

f'(t) = (3sqrt(t))/2 + 1/(2ln2(t+1))

Explanation:

It's easier to differentiate it in segments.

Let u = t^(3/2), hence, by the power rule for differentiation, (du)/(dt) = 3/2t^(1/2) = (3sqrt(t))/2

Let v = log_2(sqrt(t+1))
:.v=1/2(log_2(t+1))
:. 2v = log_2(t+1)
:. 2^(2v) = t+1
:. t = 2^(2v) - 1

Now, 2^(2v) = e^(2*ln2*v)

Hence, (dt)/(dv) = 2ln2*e^(2*ln2*v) = 2ln2*2^(2v)
:. (dv)/(dt) = 1/(2ln2*2^(2v))=1/(2ln2*(t+1)

Combining the two derivatives: (du)/(dt) and (dv)/(dt),
we get f'(t) = (3sqrt(t))/2 + 1/(2ln2(t+1))