# Find derivatives of the given functions : x= ln (1+t^2) y= t- arctgt ?

Apr 23, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{t}{2}$

#### Explanation:

When we have $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$,

then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here $x \left(t\right) = \ln \left(1 + {t}^{2}\right)$ and $\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{2 t}{1 + {t}^{2}}$

and $y \left(t\right) = t - \arctan t$ and therefore $\frac{\mathrm{dy}}{\mathrm{dt}} = 1 - \frac{1}{1 + {t}^{2}} = {t}^{2} / \left(1 + {t}^{2}\right)$

and hence $\frac{\mathrm{dy}}{\mathrm{dx}} = {t}^{2} / \left(2 t\right) = \frac{t}{2}$

Apr 23, 2018

See details below

#### Explanation:

We know that if $f \left(x\right) = \ln h \left(x\right)$, then f´(x)=(h´(x))/(h(x)) and if $g \left(x\right) = \arctan x$, then g´(x)=1/(1+x^2)

In our case, we have

$x \left(t\right) = \ln \left(1 + {t}^{2}\right)$ here $h \left(x\right) = 1 + {t}^{2}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{2 t}{1 + {t}^{2}}$

$y \left(t\right) = t - \arctan t$

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

Now if you look for $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \frac{1}{1 + {t}^{2}}}{2 \frac{t}{1 + {t}^{2}}} = \frac{t}{2}$