# DIFFERENTIATION (Text)

$By\ knowing\ the\ position\ of\ a\ body\ at\ various\ time\ intervals\ it\ is\ possible\ to\ find\ the\ rate\ at\ which\ the\ position\ of\ the\ body\ is$$changing.\ It\ is\ of \ very\ general\ interest\ to\ know\ a\ certain\ parameter\ at\ various\ instants\ of\ time\ and\ try\ to\ finding\ the\ rate$$at\ which\ it\ is\ changing.\ There\ are\ several\ real\ life\ situations\ where\ such\ a\ process\ needs\ to\ be\ carried\ out.\ For\ instance,\ people$ $maintaining\ a\ reservoir\ need\ to\ know\ when\ will\ a\ reservoir\ overflow\ knowing\ the\ depth\ of\ the\ water\ at\ several\ instances\ of\ time,$$Rocket\ Scientists\ need\ to\ compute\ the\ precise\ velocity\ with\ which\ the\ satellite\ needs\ to\ be\ shot\ out\ from\ the\ rocket$ $knowing\ the\ height\ of\ the\ rocket\ at\ various\ times.\ Financial\ institutions\ need\ to\ predict\ the\ changes\ in\ the\ value\ of\ a\ particular$ $stock\ knowing\ its\ present\ value.\ In\ these,\ and\ many\ such\ cases\ it\ is\ desirable\ to\ know\ how\ a\ particular\ parameter\ is\ changing$$with\ respect\ to\ some\ other\ parameter.$
$\color {royalblue} {Definition}:\ \hspace{20cm}$
$Suppose\ f\ is\ a\ real\ valued\ function,\ the\ function\ defined\ by\ \lim\ _{h\ \to\ 0}\ \frac{f(x\ +\ h)\ -\ f(x)}{x\ -\ h}$$is\ defined\ to\ be\ the\ derivative\ of\ f\ at\ x\ and\ is\ denoted\ by\ f\prime(x).$$f\prime(x)\ is\ denoted\ by\ \frac{d}{dx}\ (f(x))\ or\ \frac{dy}{dx}\ (where\ y\ =\ f(x))$
$\color {royalblue} {Formulae}:\ \hspace{20cm}$
$1.\ \frac{d}{dx}\ (x^n)\ =\ n\ x^{n\ -\ 1}$
$2.\ \frac{d}{dx}\ (\sqrt{x})\ =\ \frac{1}{2\ \sqrt{x}}$
$3.\ \frac{d}{dx}\ (e^x)\ =\ e^x$
$4.\ \frac{d}{dx}\ (log\ x)\ =\ \frac{1}{x}$
$5.\ \frac{d}{dx}\ (Sin\ x)\ =\ Cos\ x$
$6.\ \frac{d}{dx}\ (Cos\ x)\ =\ -\ Sin\ x$
$7.\ \frac{d}{dx}\ (Tan\ x)\ =\ Sec^2\ x$
$8.\ \frac{d}{dx}\ (Cot\ x)\ =\ -\ Cosec^2\ x$
$9.\ \frac{d}{dx}\ (Sec\ x)\ =\ Sec\ x\ Tan\ x$
$10.\ \frac{d}{dx}\ (Cosec\ x)\ =\ -\ Cosec\ x\ Cot\ x$
$11.\ \frac{d}{dx}\ (a^x)\ =\ a^x\ log\ a$
$12.\ \frac{d}{dx}\ (Sin^{-1}\ x)\ =\ \frac{1}{\sqrt{1\ -\ x^2}}$
$13.\ \frac{d}{dx}\ (Cos^{-1}\ x)\ =\ -\ \frac{1}{\sqrt{1\ -\ x^2}}$
$14.\ \frac{d}{dx}\ (Tan^{-1}\ x)\ =\ \frac{1}{1\ +\ x^2}$
$\color {royalblue} {Properties}:\ \hspace{20cm}$
$1.\ If\ u\ and\ v\ are\ functions\ of\ x,\ Then\ \frac{d}{dx}\ (u\ \pm\ v)\ =\ \frac{du}{dx}\ \pm\ \frac{dv}{dx}$
$2.\ If\ u\ is\ a\ function\ of\ x,\ and\ k\ is\ a\ constant,\ Then\ \frac{d}{dx}\ (ku)\ =\ k \frac{du}{dx}$
$3.\ If\ k\ is\ any\ constant,\ Then\ \frac{d}{dx}\ (k)\ =\ 0$
$4.\ If\ u\ and\ v\ are\ functions\ of\ x,\ Then\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}$
$4.\ If\ u\ v\ and\ w\ are\ functions\ of\ x,\ Then\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}$
$4.\ If\ u\ and\ v\ are\ functions\ of\ x,\ Then\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}$
$\color {purple} {Example\ 1:}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ x\ +\ x^2\ +\ x^3\ \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ x\ +\ x^2\ +\ x^3\ \hspace{15cm}$
$\frac{dy}{dx}\ =\ \frac{d}{dx}(x)\ +\ \frac{d}{dx}(x^2)\ +\ \frac{d}{dx}(x^3)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 1\ +\ 2\ x\ +\ 3\ x^2\ \hspace{10cm}$
$\color {purple} {Example\ 2:}\ \color {Red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ 3\ x^2\ \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ 3\ x^2\ \hspace{15cm}$
$\frac{dy}{dx}\ =\ 3\ \frac{d}{dx}(x^2)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 3\ (2x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 6\ x\ \hspace{10cm}$
$\color {purple} {Example\ 3:}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ 3\ \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ 3\ \hspace{15cm}$
$\frac{dy}{dx}\ =\ \frac{d}{dx}(3)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 0\ \hspace{10cm}$
$\color {purple} {Example\ 4:}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ x^2\ +\ x^3\ +\ Cos\ x\ \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ x^2\ +\ x^3\ +\ Cos\ x\ \hspace{15cm}$
$\frac{dy}{dx}\ =\ \frac{d}{dx}(x^2)\ +\ \frac{d}{dx}(x^3)\ +\ \frac{d}{dx}(Cos\ x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 2\ x\ +\ 3\ x^2\ -\ Sin\ x\ \hspace{10cm}$
$\color {purple} {Example\ 5:}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ \frac{5}{x^2}\ +\ \frac{2}{x}\ +\ \frac{3}{Cos\ x}\ +\ \frac{1}{8} \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ \frac{5}{x^2}\ +\ \frac{2}{x}\ +\ \frac{3}{Cos\ x}\ +\ \frac{1}{8}\ \hspace{15cm}$
$\frac{dy}{dx}\ =\ 5\ \frac{d}{dx}(x^{-2})\ +\ 2\ \frac{d}{dx}(x^{-1})\ +\ 3\ \frac{d}{dx}(Sec\ x)\ +\ \frac{d}{dx}(\frac{1}{8})\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 5\ (-\ 2\ x^{-3})\ +\ 2\ (-\ x^{-2})\ +\ 3\ Sec\ x\ Tam\ x\ +\ 0\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ -\ 10\ x^{-3}\ -\ 2\ x^{-2}\ +\ 3\ Sec\ x\ Tam\ x\ \hspace{10cm}$
$\color {purple} {Example\ 6:}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ e^x\ log\ x\ \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ e^x\ log\ x\ \hspace{15cm}$
$Here\ u\ =\ e^x,\ \hspace{5cm}\ v\ =\ log\ x$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ e^x\ \frac{d}{dx}(log\ x)\ +\ log\ x\ \frac{d}{dx}(e^x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ e^x\ \frac{1}{x}\ +\ log\ x\ e^x\ \hspace{10cm}$
$\color {purple} {Example\ 7:}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ (x\ +\ 2)\ (x\ -\ 3)\ \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ (x\ +\ 2)\ (x\ -\ 3)\ \hspace{15cm}$
$Here\ u\ =\ (x\ +\ 2),\ \hspace{5cm}\ v\ =\ (x\ -\ 3)$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ (x\ +\ 2)\ \frac{d}{dx}(x\ -\ 3)\ +\ (x\ -\ 3)\ \frac{d}{dx}(x\ +\ 2)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ (x\ +\ 2)\ (1)\ +\ (x\ -\ 3)\ (1)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ x\ +\ 2\ +\ x\ -\ 3\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 2\ x\ -\ 1\ \hspace{10cm}$
$\color {purple} {Example\ 8:}\ \color {red} {Differentiate\ the\ following\ with\ respect\ to\ x}\ (i)\ if\ y\ =\ x^3\ (1\ +\ log\ x)\ (ii)\ y\ =\ \frac{x +\ Tan\ x}{Cos\ x}\ \hspace{15cm}$
$\color {blue}{Solution:}\ (i)\ y\ =\ x^3\ (1\ +\ log\ x)\ \hspace{15cm}$
$Here\ u\ =\ x^3,\ \hspace{5cm}\ v\ =\ (1\ +\ log\ x)$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ x^3\ \frac{d}{dx}(1\ +\ log\ x)\ +\ (1\ +\ Log\ x)\ \frac{d}{dx}(x^3)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ x^3\ (0\ +\ \frac{1}{x})\ +\ (1\ +\ log\ x)\ (3\ x^2)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ x^2\ +\ 3\ x^2\ (1\ +\ log\ x)\hspace{10cm}$
$(ii)\ y\ =\ \frac{x\ +\ Tan\ x}{Cos\ x}\ \hspace{15cm}$
$Here\ u\ =\ (x\ +\ Tan\ x),\ \hspace{5cm}\ v\ =\ Cos\ x$
$W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{Cos\ x\ \frac{d}{dx}\ (x\ +\ Tan\ x)\ -\ (x\ +\ Tan\ x)\ \frac{d}{dx}\ (Cos\ x)}{(Cos\ x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{Cos\ x\ (1\ +\ Sec^2\ x)\ -\ (x\ +\ Tan\ x)\ (-Sin\ x)}{(Cos\ x)^2}\ \hspace{10cm}$

$\color {purple} {Example\ 9:}\ \color {red} {Differentiate\ the\ following\ with\ respect\ to\ x}\ (i)\ if\ y\ =\ x^3\ Sin\ x\ Tan\ x\ (ii)\ y\ =\ \frac{x +\ 6}{x\ -\ 7}\ \hspace{15cm}$
$\color {blue}{Solution:}\ (i)\ y\ =\ x^3\ Sin\ x\ Tan\ x\ \hspace{15cm}$
$Here\ u\ =\ x^3,\ \hspace{2cm}\ v\ =\ Sin\ x\ \hspace{2cm}\ w\ =\ Tan\ x$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ x^3\ Sin\ x\ \frac{d}{dx}\ (Tan\ x)\ +\ Sin\ x\ Tan\ x\ \frac{d}{dx}\ (x^3)\ +\ Tan\ x\ x^3\ \frac{d}{dx}\ (Sin\ x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ x^3\ Sin\ x\ (Sec^2\ x)\ +\ Sin\ x\ Tan\ x\ 3\ x^2\ +\ Tan\ x\ x^3\ Cos\ x\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ x^3\ Sin\ x\ (Sec^2\ x)\ +\ 3\ Sin\ x\ Tan\ x\ x^2\ +\ x^3\ Tan\ x\ Cos\ x\ \hspace{10cm}$
$(ii)\ y\ =\ \frac{x\ +\ 6}{x\ -\ 7}\ \hspace{15cm}$
$Here\ u\ =\ (x\ +\ 6),\ \hspace{5cm}\ v\ =\ (x\ -\ 7)$
$W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(x\ -\ 7)\ \frac{d}{dx}\ (x\ +\ 6)\ -\ (x\ +\ 6)\ \frac{d}{dx}\ (x\ -\ 7)}{(x\ -\ 7)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(x\ -\ 7)\ (1\ +\ 0)\ -\ (x\ +\ 6)\ (1\ -\ 0)}{(x\ -\ 7)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{x\ -\ 7\ -\ x\ -\ 6}{(x\ -\ 7)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{-\ 13}{(x\ -\ 7)^2}\ \hspace{10cm}$

$\color {purple} {Example\ 10.} \color {red} {Find \frac{dy}{dx}}\ (i)\ if\ y\ =\ (x\ +\ 1)(x\ +\ 2)(x\ +\ 3)\ \hspace{2cm}\ (ii)\ y\ =\ \frac{x^2\ +\ 1}{e^x}\ \hspace{10cm}$
$\color {black}{Solution:}\ (i)\ y\ =\ (x\ +\ 1)(x\ +\ 2)(x\ +\ 3)\ \hspace{15cm}$
$Here\ u\ =\ x\ +\ 1,\ \hspace{2cm}\ v\ =\ x\ +\ 2\ \hspace{2cm}\ w\ =\ x\ +\ 3$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ (x\ +\ 1)\ (x\ +\ 2)\ \frac{d}{dx}\ (x\ +\ 3 )\ +\ (x\ +\ 2)\ (x\ +\ 3)\ \frac{d}{dx}\ (x\ +\ 1)\ +\ (x\ +\ 3)\ (x\ +\ 1)\ \frac{d}{dx}\ (x\ +\ 3)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ (x\ +\ 1)\ (x\ +\ 2)\ (1\ +\ 0)\ +\ (x\ +\ 2)\ (x\ +\ 3)\ (1\ +\ 0)\ +\ (x\ +\ 3)\ (x\ +\ 1)\ (1\ +\ 0)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ (x\ +\ 1)\ (x\ +\ 2)\ +\ (x\ +\ 2)\ (x\ +\ 3)\ +\ (x\ +\ 3)\ (x\ +\ 1)\ \hspace{10cm}$
$(ii)\ y\ =\ \frac{x^2\ +\ 1}{e^x}\ \hspace{15cm}$
$Here\ u\ =\ (x^2\ +\ 1),\ \hspace{5cm}\ v\ =\ (e^x)$
$W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{e^x\ \frac{d}{dx}\ (x^2\ +\ 1)\ -\ (x^2\ +\ 1)\ \frac{d}{dx}\ (e^x)}{(e^x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{e^x\ (2\ x)\ (x^2\ +\ 1)\ -\ (x^2\ +\ 1)\ e^x}{(e^x)^2}\ \hspace{10cm}$

$\color {purple} {Example\ 11.:}\ \color {red} {Find\ \frac{dy}{dx}}\ (i)\ if\ y\ =\ e^x\ log\ x\ Cos\ x\ (ii)\ y\ =\ \frac{x^2\ +\ Tan\ x}{x\ -\ Sin\ x}\ \hspace{15cm}$
$\color {blue}{Solution:}\ (i)\ y\ =\ e^x\ log\ x\ Cos\ x\ \hspace{15cm}$
$Here\ u\ =\ e^x,\ \hspace{2cm}\ v\ =\ log\ x\ \hspace{2cm}\ w\ =\ Cos\ x$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ e^x\ log\ x\ \frac{d}{dx}\ (Cos\ x)\ +\ log\ x\ Cos\ x\ \frac{d}{dx}\ (e^x)\ +\ Cos\ x\ e^x\ \frac{d}{dx}\ (log\ x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ e^x\ log\ x\ (-\ Sin\ x)\ +\ log\ x\ Cos\ x\ e^x\ +\ Cos\ x\ e^x\ \frac{1}{x}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ -\ e^x\ log\ x\ Sin\ x\ +\ log\ x\ Cos\ x\ e^x\ +\ Cos\ x\ e^x\ \frac{1}{x}\ \hspace{10cm}$
$(ii)\ y\ =\ \frac{x^2\ +\ Tan\ x}{x\ -\ Sin\ x}\ \hspace{15cm}$
$Here\ u\ =\ (x^2\ +\ Tan\ x),\ \hspace{5cm}\ v\ =\ (x\ -\ Sin\ x)$
$W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(x\ -\ Sin\ x)\ \frac{d}{dx}\ (x^2\ +\ Tan\ x)\ -\ (x^2\ +\ Tan\ x)\ \frac{d}{dx}\ (x\ -\ Sin\ x)}{(x\ -\ Sin\ x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(x\ -\ Sin\ x)\ (2\ x\ +\ Sec^2\ x)\ -\ (x^2\ +\ Tan\ x)\ (1\ -\ Cos\ x)}{(x\ -\ Sin\ x)^2}\ \hspace{10cm}$
$\color {purple} {Example\ 12:}\ \color {red} {Find\ \frac{dy}{dx}}\ (i)\ if\ y\ =\ e^x\ log\ x\ \sqrt{x}\ (ii)\ y\ =\ \frac{sin\ x}{1\ +\ cos\ x}\ \hspace{15cm}$
$\color {blue}{Solution:}\ (i)\ y\ =\ e^x\ log\ x\ \sqrt{x}\ \hspace{15cm}$
$Here\ u\ =\ e^x,\ \hspace{2cm}\ v\ =\ log\ x\ \hspace{2cm}\ w\ =\ \sqrt{x}$
$W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ e^x\ log\ x\ \frac{d}{dx}\ (\sqrt{x})\ +\ log\ x\ \sqrt{x}\ \frac{d}{dx}\ (e^x)\ +\ \sqrt{x}\ e^x\ \frac{d}{dx}\ (log\ x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ e^x\ log\ x\ (\frac{1}{2\sqrt{x}})\ +\ log\ x\ \sqrt{x}\ e^x\ +\ \sqrt{x}\ e^x\ \frac{1}{x}\ \hspace{10cm}$

$(ii)\ y\ =\ \frac{sin\ x}{1\ +\ cos\ x}\ \hspace{15cm}$
$Here\ u\ =\ (Sin\ x),\ \hspace{5cm}\ v\ =\ (1\ +\ Cos\ x)$
$W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ \frac{d}{dx}\ (Sin\ x)\ -\ (Sin\ x)\ \frac{d}{dx}\ (1\ +\ Cos\ x)}{(1\ +\ Cos\ x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{(1\ -\ Cos\ x)\ (Cos\ x)\ -\ (Sin\ x)\ (-\ Sin\ x)}{(1\ +\ Cos\ x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{Cos\ x\ +\ cos^2\ x\ +\ sin^2\ x}{(1\ +\ Cos\ x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{Cos\ x\ +\ 1}{(1\ +\ Cos\ x)^2}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{1}{(1\ +\ Cos\ x)}\ \hspace{10cm}$

## Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1.}\ \color {red} {Find}\ \frac{dy}{dx}\ if\ y\ =\ \frac{1}{x^2} +\ 3\ tan\ x\ -\ log\ x\ \hspace{15cm}$
$\color {purple} {2.}\ \color {red} {Find}\ \frac{dy}{dx}\ if\ y\ =\ 7\ e^x\ \ +\ 4\ log\ x\ +\ \frac{1}{x^2}\ \hspace{15cm}$
$\color {purple} {3.} \ \color {red} {Find}\ \frac{dy}{dx}\ if\ y\ =\ 8\ e^x\ -\ 4\ cosec\ x\ \hspace{15cm}$
$\color {purple} {4.}\ \color {red} {Find}\ \frac{dy}{dx}\ if\ y\ =\ e^x\ Cos\ x\ \hspace{15cm}$
$\color {purple} {5.}\ \color {red} {Find}\ \frac{dy}{dx}\ if\ y\ =\ x^4\ Sin\ x\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {6.}\ \color {red} {Find}\ \frac{dy}{dx}\ if\ y\ =\ x^3\ log\ x\ \hspace{15cm}$
$\color {purple} {7.}\ If\ y\ =\ (x\ +\ 3)\ (x\ -\ 4)\ \color {red} {find\ \frac{dy}{dx}}\ \hspace{15cm}$
$\color {purple} {8.}\ \color {red} {Find}\ \frac{dy}{dx}\ if\ y\ =\ x^2\ e^x\ Sin\ x\ \hspace{15cm}$
$\color {purple} {9.}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ (x^2\ +\ 5)\ Cos\ x\ e^{-\ 2x}\ \hspace{15cm}$
$\color {purple} {10.}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ \frac{Sin\ x}{log\ x}\ \hspace{15cm}$
$\color {purple} {11}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ \frac{1\ +\ cos\ x}{1\ -\ cos\ x}\ \hspace{15cm}$
$\color {purple} {12.}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ y\ =\ \frac{x\ Sin\ x}{e^x}\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {13.}\ \color {red} {Find}\ \frac{dy}{dx}\ (i)\ if\ y\ =\ x\ e^x\ log\ x\ (ii)\ y\ =\ \ (x^2\ +\ 2)\ Cos\ x\ \hspace{15cm}$
$\color {purple} {14.}\ \color {red} {Find}\ \frac{dy}{dx}\ (i)\ if\ y\ =\ e^x\ x\ Cos\ x\ (ii)\ y\ =\ \frac{x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{15cm}$
$\color {purple} {15.}\ \color {red} {Find}\ \frac{dy}{dx}\ (i)\ if\ y\ =\ (2\ x\ +\ 1)(3\ x\ -\ 7)(4\ -\ 9\ x)\ \hspace{2cm}\ (ii)\ y\ =\ \frac{e^x\ +\ Sin\ x}{1\ -\ Cos\ x}\ \hspace{10cm}$
$\color {purple} {16:}\ \color {red} {Find\ \frac{dy}{dx}}\ (i)\ if\ y\ =\ log\ x\ (2x\ +\ 1)\ \sqrt{x} \ \hspace{2cm}\ (ii)\ y\ =\ \frac{ax\ +\ b}{cx\ +\ d}\ \hspace{15cm}$