# Differentiation (Revision)

$1:\ Find\ \frac{dy}{dx}\ (i)\ if\ y\ =\ e^x\ x\ Cos\ x\ \hspace{2cm}\ (ii)\ y\ =\ \frac{ x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{5cm}$
$\color {black}{Solution:}\ (i)\ y\ =\ e^x\ x\ Cos\ x\ \hspace{15cm}$
$Here\ u\ =\ e^x\ \hspace{2cm}\ v\ =\ x\ \hspace{2cm}\ w\ =\ Cos\ x$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ e^x\ x\ \frac{d}{dx}\ (Cos\ x)\ +\ x\ Cos\ x\ \frac{d}{dx}\ (e^x)\ +\ Cos\ x\ e^x\ \frac{d}{dx}\ (x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ e^x\ x\ (-\ Sin\ x)\ +\ x\ Cos\ x\ e^x\ +\ Cos\ x\ e^x\ (1)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ -\ e^x\ x\ Sin\ x\ +\ x\ Cos\ x\ e^x\ +\ Cos\ x\ e^x\ \hspace{10cm}$
$(ii)\ y\ =\ \frac{ x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{15cm}$
$Here\ u\ =\ (x\ +\ Sin\ x),\ \hspace{5cm}\ v\ =\ (1\ -\ Cos\ x)$
$W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ \frac{d}{dx}\ (x\ +\ Sin\ x)\ -\ (x\ +\ Sin\ x)\ \frac{d}{dx}\ (1\ -\ Cos\ x)}{(1\ -\ Cos\ x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ (1\ +\ Cos\ x)\ -\ (x\ +\ Sin\ x)\ (Sin\ x)}{(x\ -\ Sin\ x)^2}\ \hspace{10cm}$
$2:\ For\ (i)\ if\ y\ =\ \frac{ e^x\ +\ Sin\ x}{1\ +\ Tan\ x}\ \hspace{2cm}\ (ii)\ y\ =\ (x\ +\ Sin\ x)(1\ +\ Cos\ x)\ ,\ find\ \frac{dy}{dx}\ \hspace{5cm}$
$\color {black}{Solution:}\ (i)\ y\ =\ \frac{ e^x\ +\ Sin\ x}{1\ +\ Tan\ x}\ \hspace{15cm}$
$Here\ u\ =\ (e^x\ +\ Sin\ x),\ \hspace{5cm}\ v\ =\ (1\ +\ Tan\ x)$
$W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(1\ +\ Tan\ x)\ \frac{d}{dx}\ (e^x\ +\ Sin\ x)\ -\ (e^x\ +\ Sin\ x)\ \frac{d}{dx}\ (1\ +\ Tan\ x)}{(1\ +\ Tan\ x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(1\ +\ Tan\ x)\ (e^x\ +\ Cos\ x)\ -\ (e^x\ +\ Sin\ x)\ (Sec^2\ x)}{(1\ +\ Tan\ x)^2}\ \hspace{10cm}$
$(ii)\ y\ =\ (x\ +\ Sin\ x)(1\ +\ Cos\ x)\ \hspace{15cm}$
$Here\ u\ =\ (x\ +\ Sin\ x),\ \hspace{5cm}\ v\ =\ (1\ +\ Cos\ x)$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ (x\ +\ Sin\ x)\ \frac{d}{dx}(1\ +\ Cos\ x)\ +\ (1\ +\ Cos\ x)\ \frac{d}{dx}(x\ +\ Sin\ x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ (x\ +\ Sin\ x)\ (-\ Sin\ x)\ +\ (1\ +\ Cos\ x)\ (1\ +\ Cos\ x)\ \hspace{10cm}$