# How do you find the derivative of f(t)=t^(3/2)log_2sqrt(t+1)?

Dec 23, 2017

$f ' \left(t\right) = {t}^{\frac{3}{2}} \left(\frac{1}{2 \ln 2 \left(t + 1\right)}\right) + {\log}_{2} \sqrt{t + 1} \left(\frac{3}{2} {t}^{\frac{1}{2}}\right)$

#### Explanation:

.

$f \left(t\right) = {t}^{\frac{3}{2}} {\log}_{2} \sqrt{t + 1}$

This is a product of two functions. As such, we use the product rule which says if:

$f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$

$f ' \left(x\right) = g \left(x\right) \cdot h ' \left(x\right) + h \left(x\right) \cdot g ' \left(x\right)$

In our problem,

$f \left(t\right) = g \left(t\right) \cdot h \left(t\right)$

We need to figure out:

$f ' \left(t\right) = g \left(t\right) \cdot h ' \left(t\right) + h \left(t\right) \cdot g ' \left(t\right)$

$g \left(t\right) = {t}^{\frac{3}{2}}$

$h \left(t\right) = {\log}_{2} \sqrt{t + 1}$

$g ' \left(t\right) = \frac{3}{2} {t}^{\frac{1}{2}}$

For $h ' \left(t\right)$, we can say $u = \sqrt{t + 1} = {\left(t + 1\right)}^{\frac{1}{2}}$

du=1/2(t+1)^(-1/2)=1/(2(t+1)^(1/2))=1/(2sqrt(t+1)

Then. we can say:

$h \left(t\right) = {\log}_{2} u$

$h ' \left(t\right) = \left(\frac{1}{u \left(\ln 2\right)}\right) \mathrm{du}$

h'(t)=(1/(sqrt(t+1)(ln2)))(1/(2sqrt(t+1)))=1/(2ln2(t+1)

Now, we can plug these results in:

$f ' \left(t\right) = {t}^{\frac{3}{2}} \left(\frac{1}{2 \ln 2 \left(t + 1\right)}\right) + {\log}_{2} \sqrt{t + 1} \left(\frac{3}{2} {t}^{\frac{1}{2}}\right)$