# How do you find the second derivative of f(t)=tsqrtt?

Jun 21, 2017

$\frac{{d}^{2} f}{{\mathrm{dt}}^{2}} = \frac{3}{4 \sqrt{t}}$

#### Explanation:

As $t \sqrt{t}$ can be written as $f \left(t\right) = t \sqrt{t} = t \times {t}^{\frac{1}{2}} = {t}^{\frac{3}{2}}$, we can use the formula $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

Hence $\frac{\mathrm{df}}{\mathrm{dt}} = \frac{3}{2} \times {t}^{\frac{3}{2} - 1} = \frac{3}{2} {t}^{\frac{1}{2}} = \frac{3}{2} \sqrt{t}$

and $\frac{{d}^{2} f}{{\mathrm{dt}}^{2}} = \frac{3}{2} \times \frac{1}{2} \times {t}^{\frac{1}{2} - 1} = \frac{3}{4} {t}^{- \frac{1}{2}} = \frac{3}{4 \sqrt{t}}$

Jun 21, 2017

$f ' ' \left(t\right) = \frac{3}{4 \sqrt{t}}$

#### Explanation:

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

$\text{differentiate using the "color(blue)"power rule}$

• d/dx(ax^n)=nax^(n-1)

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

$\Rightarrow f ' ' \left(t\right) = \frac{3}{2} \times \frac{1}{2} {t}^{- \frac{1}{2}}$

$\textcolor{w h i t e}{\Rightarrow f ' ' \left(t\right)} = \frac{3}{4} \times \frac{1}{t} ^ \left(\frac{1}{2}\right) = \frac{3}{4 \sqrt{t}}$