# What is the derivative of y = ln(t^2 + 4) - 1/2 arctan(t/2)?

## What is the derivative of $y = \ln \left({t}^{2} + 4\right) - \frac{1}{2} \arctan \left(\frac{t}{2}\right)$ with respec to $t$?

Jul 19, 2018

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2 t - 1}{{t}^{2} + 4}$.

#### Explanation:

$y = \ln \left({t}^{2} + 4\right) - \frac{1}{2} \arctan \left(\frac{t}{2}\right)$.

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left\{\ln \left({t}^{2} + 4\right) - \frac{1}{2} \arctan \left(\frac{t}{2}\right)\right\}$,

$= \frac{d}{\mathrm{dt}} \left\{\ln \left({t}^{2} + 4\right)\right\} - \frac{1}{2} \frac{d}{\mathrm{dt}} \left\{\arctan \left(\frac{t}{2}\right)\right\}$,

$= \frac{1}{{t}^{2} + 4} \cdot \frac{d}{\mathrm{dt}} \left({t}^{2} + 4\right) - \frac{1}{2} \cdot \frac{1}{1 + {\left(\frac{t}{2}\right)}^{2}} \cdot \frac{d}{\mathrm{dt}} \left\{\frac{t}{2}\right\}$,

$= \frac{1}{{t}^{2} + 4} \cdot \left(2 t\right) - \frac{1}{2} \cdot \frac{1}{1 + {t}^{2} / 4} \cdot \frac{1}{2}$,

$= \frac{2 t}{{t}^{2} + 4} - \frac{1}{4} \cdot \frac{4}{4 + {t}^{2}}$.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2 t - 1}{{t}^{2} + 4}$, as desired!

Jul 19, 2018

$f ' \left(t\right) = \frac{2 t - 1}{{t}^{2} + 4}$

#### Explanation:

Note that

$\left(\ln \left(x\right)\right) ' = \frac{1}{x}$ and $\left(\arctan \left(x\right)\right) ' = \frac{1}{1 + {x}^{2}}$
and additionally we use the chain rule

$\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
so we get

$f ' \left(t\right) = \frac{2 t}{{t}^{2} + 4} - \frac{1}{2} \cdot \frac{1}{1 + {\left(\frac{t}{2}\right)}^{2}} \cdot \frac{1}{2}$

note that

$\frac{1}{4} \cdot \frac{1}{1 + {t}^{2} / 4} = \frac{1}{4} \cdot \frac{4}{4 + {t}^{2}} = \frac{1}{4 + {t}^{2}}$
so we get

$y ' = \frac{2 t - 1}{{t}^{2} + 4}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2 t - 1}{{t}^{2} + 4}$

#### Explanation:

Given that

$y = \setminus \ln \left({t}^{2} + 4\right) - \frac{1}{2} \setminus {\tan}^{- 1} \left(\frac{t}{2}\right)$

differentiating above function w.r.t. $t$ using chain rule as follows

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(\ln \left({t}^{2} + 4\right) - \frac{1}{2} \setminus {\tan}^{- 1} \left(\frac{t}{2}\right)\right)$

$= \frac{1}{{t}^{2} + 4} \frac{d}{\mathrm{dt}} \left({t}^{2} + 4\right) - \frac{1}{2} \setminus \frac{1}{1 + {\left(\frac{t}{2}\right)}^{2}} \frac{d}{\mathrm{dt}} \left(\frac{t}{2}\right)$

$= \frac{1}{{t}^{2} + 4} \left(2 t\right) - \frac{1}{2} \setminus \frac{4}{{t}^{2} + 4} \left(\frac{1}{2}\right)$

$= \frac{2 t}{{t}^{2} + 4} - \setminus \frac{1}{{t}^{2} + 4}$

$= \frac{2 t - 1}{{t}^{2} + 4}$