DIFFERENTIATION

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