DIFFERENTIATION (Excercise)

\[\underline{PART\ -\ A}\]
\[1.\ Find\ \frac{dy}{dx}\ if\ y\ =\ 8\ e^x\ -\ 4\ cosec\ x\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ 8\ e^x\ -\ 4\ cosec\ x\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ 8\ \frac{d}{dx}(e^x)\ +\ \frac{d}{dx}(x^3)\ +\ \frac{d}{dx}(Cos\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 8\ e^x\ +\ 3\ x^2\ -\ Sin\ x\ \hspace{10cm}\]
\[\underline{PART\ -\ B}\]
\[2.\ Find\ \frac{dy}{dx}\ if\ y\ =\ (x^2\ +\ 5)\ Cos\ x\ e^{-\ 2x}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ (x^2\ +\ 5)\ Cos\ x\ e^{-\ 2x}\ \hspace{15cm}\]
\[Here\ u\ =\ x^2\ +\ 5,\ \hspace{2cm}\ v\ =\ Cos\ x\ \hspace{2cm}\ w\ =\ e^{-\ 2x}\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ (x^2\ +\ 5)\ Cos\ x\ \frac{d}{dx}\ (e^{-\ 2x})\ +\ Cos\ x\ e^{-\ 2x}\ \frac{d}{dx}\ (x^2\ +\ 5)\ +\ e^{-\ 2x}\ (x^2\ +\ 5)\ \frac{d}{dx}\ (Cos\ x)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ (x^2\ +\ 5)\ Cos\ x\ (-\ 2\ e^{-\ 2x})\ +\ Cos\ x\ e^{-\ 2x}\ (2\ x)\ +\ e^{-\ 2x}\ (x^2\ +\ 5)\ (-\ Sin\ x)\ \hspace{10cm}\]
\[\underline{PART\ -\ C}\]
\[3.\ \frac{dy}{dx}\ (i)\ if\ y\ =\ (2\ x\ +\ 1)(3\ x\ -\ 7)(4\ -\ 9\ x)\ \hspace{2cm}\ (ii)\ y\ =\ \frac{e^x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ (i)\ y\ =\ (2\ x\ +\ 1)(3\ x\ -\ 7)(4\ -\ 9\ x)\ \hspace{15cm}\]
\[Here\ u\ =\ 2\ x\ +\ 1,\ \hspace{2cm}\ v\ =\ 3\ x\ -\ 7\ \hspace{2cm}\ w\ =\ 4\ -\ 9\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ (2\ x\ +\ 1)\ (3\ x\ -\ 7)\ \frac{d}{dx}\ (4\ -\ 9\ x)\ +\ (3\ x\ -\ 7)\ (4\ -\ 9\ x)\ \frac{d}{dx}\ (2\ x\ +\ 1)\ +\ (4\ -\ 9\ x)\ (2\ x\ +\ 1)\ \frac{d}{dx}\ (3\ x\ -\ 7)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ (2\ x\ +\ 1)\ (3\ x\ -\ 7)\ (-\ 9(1))\ +\ (3\ x\ -\ 7)\ (4\ -\ 9\ x)\ 2(1)\ +\ (4\ -\ 9\ x)\ (2\ x\ +\ 1)\ 3(1)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ -\ 9(2\ x\ +\ 1)\ (3\ x\ -\ 7)\ +\ 2\ (3\ x\ -\ 7)\ (4\ -\ 9\ x)\ 2(1)\ +\ 3\ (4\ -\ 9\ x)\ (2\ x\ +\ 1)\ \hspace{10cm}\]
\[(ii)\ y\ =\ \frac{e^x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{15cm}\]
\[Here\ u\ =\ (e^x\ +\ Sin\ x),\ \hspace{5cm}\ v\ =\ (1\ -\ Cos\ x)\]
\[W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ \frac{d}{dx}\ (e^x\ +\ Sin\ x)\ -\ (e^x\ +\ Sin\ x)\ \frac{d}{dx}\ (1\ -\ Cos\ x)}{(1\ -\ Cos\ x)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ (e^x\ +\ Cos\ x)\ -\ (e^x\ +\ Sin\ x)\ (Sin\ x)}{(1\ -\ Cos\ x)^2}\ \hspace{10cm}\]