DIFFERENTIATION (Excercise)

\[\underline{PART\ -\ A}\]
\[1.\ Find\ \frac{dy}{dx}\ if\ y\ =\ \frac{1}{x^2} +\ 3\ tan\ x\ -\ log\ x\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ \frac{1}{x^2} +\ 3\ tan\ x\ -\ log\ x\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ \frac{d}{dx}(x^{-2})\ +\ 3\ \frac{d}{dx}(tan\ x)\ -\ \frac{d}{dx}(log\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ (-\ 2)\ x^{-2\ -\ 1}\ +\ 3\ Sec^2\ x\ -\ \frac{1}{x}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ -\ 2\ x^{-3}\ +\ 3\ Sec^2\ x\ -\ \frac{1}{x}\ \hspace{10cm}\]
\[2.\ Find\ \frac{dy}{dx}\ if\ y\ =\ 7\ e^x\ \ +\ 4\ log\ x\ +\ \frac{1}{x^2}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ 7\ e^x\ \ +\ 4\ log\ x\ +\ \frac{1}{x^2}\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ 7\ \frac{d}{dx}(e^x)\ +\ 4\ \frac{d}{dx}(log\ x)\ +\ \frac{d}{dx}(x^{-2})\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 7\ e^x\ +\ 4\ \frac{1}{x}\ +\ (-\ 2)\ x^{-2\ -\ 1}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 7\ e^x\ +\ \frac{4}{x}\ -\ 2\ x^{-3}\ \hspace{10cm}\]
\[3.\ Find\ \frac{dy}{dx}\ if\ y\ =\ 8\ e^x\ -\ 4\ cosec\ x\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ 8\ e^x\ -\ 4\ cosec\ x\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ 8\ \frac{d}{dx}(e^x)\ +\ \frac{d}{dx}(x^3)\ +\ \frac{d}{dx}(Cos\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 8\ e^x\ +\ 3\ x^2\ -\ Sin\ x\ \hspace{10cm}\]
\[4.\ Find\ \frac{dy}{dx}\ if\ y\ =\ e^x\ Cos\ x\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ e^x\ Cos\ x\ \hspace{15cm}\]
\[Here\ u\ =\ e^x,\ \hspace{5cm}\ v\ =\ Cos\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ e^x\ \frac{d}{dx}(Cos\ x)\ +\ Cos\ x\ \frac{d}{dx}(e^x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ e^x\ \frac{1}{x}\ +\ Cos\ x\ e^x\ \hspace{10cm}\]
\[5.\ Find\ \frac{dy}{dx}\ if\ y\ =\ x^4\ Sin\ x\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ x^4\ Sin\ x\ \hspace{15cm}\]
\[Here\ u\ =\ x^4,\ \hspace{5cm}\ v\ =\ Sin\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x^4\ \frac{d}{dx}(Sin\ x)\ +\ Sin\ x\ \frac{d}{dx}(x^4)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x^4\ Cos\ x\ +\ Sin\ x\ 4(x^3)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x^4\ Cos\ x\ +\ 4\ Sin\ x\ x^3\ \hspace{10cm}\]
\[\underline{PART\ -\ B}\]
\[6.\ If\ y\ =\ (x\ +\ 3)\ (x\ -\ 4)\ find\ \frac{dy}{dx}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ (x\ +\ 3)\ (x\ -\ 4)\ \hspace{15cm}\]
\[Here\ u\ =\ (x\ +\ 3),\ \hspace{5cm}\ v\ =\ (x\ -\ 4)\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ (x\ +\ 3)\ \frac{d}{dx}(x\ -\ 4)\ +\ (x\ -\ 4)\ \frac{d}{dx}(x\ +\ 3)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ (x\ +\ 3)\ (1)\ +\ (x\ -\ 4)\ (1)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x\ +\ 3\ +\ x\ -\ 4\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 2\ x\ -\ 1\ \hspace{10cm}\]
\[7.\ Find\ \frac{dy}{dx}\ if\ y\ =\ x^2\ e^x\ Sin\ x\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ x^2\ e^x\ Sin\ x\ \hspace{15cm}\]
\[Here\ u\ =\ x^2,\ \hspace{2cm}\ v\ =\ e^x\ \hspace{2cm}\ w\ =\ Sin\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ x^2\ e^x\ \frac{d}{dx}\ (Sin\ x)\ +\ e^x\ Sin\ x\ \frac{d}{dx}\ (x^2)\ +\ Sin\ x\ x^2\ \frac{d}{dx}\ (e^x)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ x^2\ e^x\ Cos\ x\ +\ e^x\ Sin\ x\ (2\ x) +\ Sin\ x\ x^2\ e^x\ \hspace{10cm}\]
\[8.\ Find\ \frac{dy}{dx}\ if\ y\ =\ (x^2\ +\ 5)\ Cos\ x\ e^{-\ 2x}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ (x^2\ +\ 5)\ Cos\ x\ e^{-\ 2x}\ \hspace{15cm}\]
\[Here\ u\ =\ x^2\ +\ 5,\ \hspace{2cm}\ v\ =\ Cos\ x\ \hspace{2cm}\ w\ =\ e^{-\ 2x}\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ (x^2\ +\ 5)\ Cos\ x\ \frac{d}{dx}\ (e^{-\ 2x})\ +\ Cos\ x\ e^{-\ 2x}\ \frac{d}{dx}\ (x^2\ +\ 5)\ +\ e^{-\ 2x}\ (x^2\ +\ 5)\ \frac{d}{dx}\ (Cos\ x)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ (x^2\ +\ 5)\ Cos\ x\ (-\ 2\ e^{-\ 2x})\ +\ Cos\ x\ e^{-\ 2x}\ (2\ x)\ +\ e^{-\ 2x}\ (x^2\ +\ 5)\ (-\ Sin\ x)\ \hspace{10cm}\]
\[9.\ Find\ \frac{dy}{dx}\ if\ y\ =\ \frac{Sin\ x}{log\ x}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ \frac{Sin\ x}{log\ x}\ \hspace{15cm}\]
\[Here\ u\ =\ Sin\ x,\ \hspace{5cm}\ v\ =\ log\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{log\ x\ \frac{d}{dx}\ (Sin\ x)\ -\ (Sin\ x)\ \frac{d}{dx}\ (log\ x)}{(log\ x)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(log\ x)\ (Cos\ x)\ -\ (Sin\ x)\ (\frac{1}{x})}{(log\ x)^2}\ \hspace{10cm}\]
\[10.\ Find\ \frac{dy}{dx}\ if\ y\ =\ \frac{1\ +\ cos\ x}{1\ -\ cos\ x}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ \frac{1\ +\ cos\ x}{1\ -\ cos\ x}\ \hspace{15cm}\]
\[Here\ u\ =\ 1\ +\ cos\ x,\ \hspace{5cm}\ v\ =\ 1\ -\ cos\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(1\ -\ cos\ x)\ \frac{d}{dx}\ (1\ +\ cos\ x)\ -\ (1\ +\ cos\ x)\ \frac{d}{dx}\ (1\ -\ cos\ x)}{(1\ -\ cos\ x)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(1\ -\ cos\ x)\ (-\ Sin\ x)\ -\ (1\ +\ cos\ x)\ (Sin\ x)}{(1\ -\ cos\ x)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{-\ Sin\ x)\ +\ Sin\ x\ Cos\ x\ -\ Sin\ x\ -\ Sin\ x\ Cos\ x}{(1\ -\ cos\ x)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{-\ 2\ Sin\ x}{(1\ -\ cos\ x)^2}\ \hspace{10cm}\]
\[11.\ Find\ \frac{dy}{dx}\ if\ y\ =\ \frac{x\ Sin\ x}{e^x}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ \frac{x\ Sin\ x}{e^x}\ \hspace{15cm}\]
\[Here\ u\ =\ x\ Sin\ x,\ \hspace{5cm}\ v\ =\ e^x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{e^x\ \frac{d}{dx}\ (x\ Sin\ x)\ -\ (Sin\ x)\ \frac{d}{dx}\ (e^x)}{(e^x)^2}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{(e^x)\ (x\ Cos\ x)\ -\ (Sin\ x)\ (e^x)}{(e^x)^2}\ \hspace{10cm}\]
\[\underline{PART\ -\ C}\]
\[12.\ Find\ \frac{dy}{dx}\ (i)\ if\ y\ =\ x\ e^x\ log\ x\ (ii)\ y\ =\ \ (x^2\ +\ 2)\ Cos\ x\ \hspace{15cm}\]
\[\color {black}{Solution:}\ (i)\ y\ =\ x\ e^x\ log\ x\ \hspace{15cm}\]
\[Here\ u\ =\ x,\ \hspace{2cm}\ v\ =\ e^x\ \hspace{2cm}\ w\ =\ log\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ x\ e^x\ \frac{d}{dx}\ (log\ x)\ +\ e^x\ log\ x\ \frac{d}{dx}\ (x)\ +\ log\ x\ x\ \frac{d}{dx}\ (e^x)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ x\ e^x\ (\frac{1}{x})\ +\ e^x\ log\ x\ (1)\ +\ log\ x\ x\ (e^x)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{x\ e^x}{x}\ +\ e^x\ log\ x\ +\ x\ log\ x\ e^x\ \hspace{10cm}\]
\[(ii)\ y\ =\ (x^2\ +\ 2)\ Cos\ x\ \hspace{15cm}\]
\[Here\ u\ =\ (x^2\ +\ 2),\ \hspace{5cm}\ v\ =\ Cos\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ (x^2\ +\ 2)\ \frac{d}{dx}(Cos\ x)\ +\ Cos\ x\ \frac{d}{dx}(x^2\ +\ 2)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ (x^2\ +\ 2)\ (-\ Sin\ x)\ +\ Cos\ x\ (2\ x\ +\ 0)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ -(x^2\ +2)\ Sin\ x\ +\ 2\ x\ Cos\ x\hspace{10cm}\]
\[13.\ Find\ \frac{dy}{dx}\ (i)\ if\ y\ =\ e^x\ x\ Cos\ x\ (ii)\ y\ =\ \frac{x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ (i)\ y\ =\ e^x\ x\ Cos\ x\ \hspace{15cm}\]
\[Here\ u\ =\ e^x,\ \hspace{2cm}\ v\ =\ x\ \hspace{2cm}\ w\ =\ Cos\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ e^x\ x\ \frac{d}{dx}\ (Cos\ x)\ +\ x\ Cos\ x\ \frac{d}{dx}\ (e^x)\ +\ Cos\ x\ e^x\ \frac{d}{dx}\ (x)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ e^x\ x\ (-\ Sin\ x)\ +\ x\ Cos\ x\ e^x\ +\ Cos\ x\ e^x\ (1)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ -\ e^x\ x\ Sin\ x\ +\ x\ Cos\ x\ e^x\ +\ Cos\ x\ e^x\ \hspace{10cm}\]
\[(ii)\ y\ =\ \frac{x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{15cm}\]
\[Here\ u\ =\ (x\ +\ Sin\ x),\ \hspace{5cm}\ v\ =\ (1\ -\ Cos\ x)\]
\[W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ \frac{d}{dx}\ (x\ +\ Sin\ x)\ -\ (x\ +\ Sin\ x)\ \frac{d}{dx}\ (1\ -\ Cos\ x)}{(1\ -\ Cos\ x)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ (1\ +\ Cos\ x)\ -\ (x\ +\ Sin\ x)\ (Sin\ x)}{(1\ -\ Cos\ x)^2}\ \hspace{10cm}\]
\[14.\ Find\ \frac{dy}{dx}\ (i)\ if\ y\ =\ e^x\ log\ x\ Cos\ x\ (ii)\ y\ =\ \frac{x^2\ +\ Tan\ x}{x\ -\ Sin\ x}\ \hspace{15cm}\]
\[15.\ \frac{dy}{dx}\ (i)\ if\ y\ =\ (2\ x\ +\ 1)(3\ x\ -\ 7)(4\ -\ 9\ x)\ \hspace{2cm}\ (ii)\ y\ =\ \frac{e^x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ (i)\ y\ =\ (2\ x\ +\ 1)(3\ x\ -\ 7)(4\ -\ 9\ x)\ \hspace{15cm}\]
\[Here\ u\ =\ 2\ x\ +\ 1,\ \hspace{2cm}\ v\ =\ 3\ x\ -\ 7\ \hspace{2cm}\ w\ =\ 4\ -\ 9\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ (2\ x\ +\ 1)\ (3\ x\ -\ 7)\ \frac{d}{dx}\ (4\ -\ 9\ x)\ +\ (3\ x\ -\ 7)\ (4\ -\ 9\ x)\ \frac{d}{dx}\ (2\ x\ +\ 1)\ +\ (4\ -\ 9\ x)\ (2\ x\ +\ 1)\ \frac{d}{dx}\ (3\ x\ -\ 7)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ (2\ x\ +\ 1)\ (3\ x\ -\ 7)\ (-\ 9(1))\ +\ (3\ x\ -\ 7)\ (4\ -\ 9\ x)\ 2(1)\ +\ (4\ -\ 9\ x)\ (2\ x\ +\ 1)\ 3(1)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ -\ 9(2\ x\ +\ 1)\ (3\ x\ -\ 7)\ +\ 2\ (3\ x\ -\ 7)\ (4\ -\ 9\ x)\ 2(1)\ +\ 3\ (4\ -\ 9\ x)\ (2\ x\ +\ 1)\ \hspace{10cm}\]
\[(ii)\ y\ =\ \frac{e^x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{15cm}\]
\[Here\ u\ =\ (e^x\ +\ Sin\ x),\ \hspace{5cm}\ v\ =\ (1\ -\ Cos\ x)\]
\[W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ \frac{d}{dx}\ (e^x\ +\ Sin\ x)\ -\ (e^x\ +\ Sin\ x)\ \frac{d}{dx}\ (1\ -\ Cos\ x)}{(1\ -\ Cos\ x)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ (e^x\ +\ Cos\ x)\ -\ (e^x\ +\ Sin\ x)\ (Sin\ x)}{(1\ -\ Cos\ x)^2}\ \hspace{10cm}\]