Question

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

Answer 1

`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))`