DIFFERENTIATION METHODS

\[\color {royalblue} {Chain Rule}:\ \hspace{20cm}\]
\[If\ ‘y’\ is\ a\ function\ of\ ‘u’\ and\ ‘u’\ is\ a\ function\ of\ ‘x’,\ Then\ \frac{dy}{dx}\ =\ \frac{dy}{du}\ ×\ \frac{du}{dx}\]
\[\color {royalblue} {Examples}:\ \hspace{20cm}\]
\[Ex\ 1:\ Find\ \frac{dy}{dx}\ if\ y\ =\ (3\ x\ +\ 6)^5\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ (3\ x\ +\ 6)^5\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ 5\ (3\ x\ +\ 6)^{5\ -\ 1}\ \frac{d}{dx}(3\ x\ +\ 6)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 5\ (3\ x\ +\ 6)^4\ 3\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 15\ (3\ x\ +\ 6)^4\ \hspace{10cm}\]
\[Ex\ 2:\ Find\ \frac{dy}{dx}\ if\ y\ =\ Sin(e^x)\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ Sin(e^x)\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ Cos(e^x)\ \frac{d}{dx}(e^x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ Cos(e^x)\ e^x\ \hspace{10cm}\]
\[Ex\ 3:\ If\ y\ =\ log(Sin\ x),\ find\ \frac{dy}{dx}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ log(Sin\ x)\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Sin\ x}\ \frac{d}{dx}(Sin\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Sin\ x}\ Cos\ x\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ Cot\ x\ \hspace{10cm}\]
\[Ex\ 4:\ Find\ \frac{dy}{dx}\ if\ y\ =\ Sin(log\ x)\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ Sin(log\ x)\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ Cos(log\ x)\ \frac{d}{dx}(log\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ Cos(log\ x)\ \frac{1}{x}\ \hspace{10cm}\]
\[Ex\ 5:\ find\ \frac{dy}{dx},\ if\ y\ =\ log(Sec\ x\ +\ Tan\ x)\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ log(Sec\ x\ +\ Tan\ x)\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Sec\ x\ +\ Tan\ x}\ \frac{d}{dx}(Sec\ x\ +\ Tan\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Sec\ x\ +\ Tan\ x}\ (Sec\ x\ Tan\ x\ +\ Sec^2\ x)\ \hspace{10cm}\]
\[Ex\ 6:\ Find\ \frac{dy}{dx}\ if\ y\ =\ Sin(e^x\ log\ x)\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ Sin(e^x\ log\ x)\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ Cos(e^x\ log\ x)\ \frac{d}{dx}(e^x\ log\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ Cos(e^x\ log\ x)\ (e^x\ \frac{d}{dx}(log\ x)\ +\ log\ x\ \frac{d}{dx}(e^x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ Cos(e^x\ log\ x)\ (e^x\ \frac{1}{x}\ +\ log\ x\ e^x)\ \hspace{10cm}\]
\[Ex\ 7:\ Find\ \frac{dy}{dx}\ if\ y\ =\ e^{3x}\ Cos^2\ x\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ e^{3x}\ Cos^2\ x\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ e^{3x}\ \frac{d}{dx}(Cos^2\ x)\ +\ Cos^2\ x\ \frac{d}{dx}(e^{3x})\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ e^{3x}\ 2\ Cos\ x\ \frac{d}{dx}(Cos\ x)\ +\ Cos^2\ x\ e^{3x}\ \frac{d}{dx}(3x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 2\ e^{3x}\ Cos\ x\ (-\ Sin\ x)\ +\ Cos^2\ x\ e^{3x}\ 3\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ -\ 2\ e^{3x}\ Cos\ x\ Sin\ x\ +\ 3\ Cos^2\ x\ e^{3x}\ \hspace{10cm}\]
\[Ex\ 8:\ If\ y\ =\ Sin^{-1}\ (3\ x),\ find\ \frac{dy}{dx}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ Sin^{-1}\ (3\ x)\ \hspace{15cm}\]
\[W.\ K.\ T\ \frac{d}{dx}\ (Sin^{-1}\ x)\ =\ \frac{1}{\sqrt{1\ -\ x^2}}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{d}{dx}((Sin^{-1}\ (3\ x))\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{\sqrt{1\ -\ (3\ x)^2}}\ \frac{d}{dx}(3\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{\sqrt{1\ -\ 9\ x^2}}\ 3\ \ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{3}{\sqrt{1\ -\ 9\ x^2}}\ \hspace{10cm}\]
\[Ex\ 9:\ If\ y\ =\ Tan^{-1}\ (x^3),\ find\ \frac{dy}{dx}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ Tan^{-1}\ (x^3)\ \hspace{15cm}\]
\[W.\ K.\ T\ \frac{d}{dx}\ (Tan^{-1}\ x)\ =\ \frac{1}{1\ +\ x^2}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{d}{dx}(Tan^{-1}\ (x^3))\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{1\ +\ (x^3)^2}\ \frac{d}{dx}(x^3)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{1\ +\ x^6}\ 3\ x^2\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{3\ x^2}{1\ +\ x^6}\ \hspace{10cm}\]
\[Ex\ 10:\ Find\ \frac{dy}{dx}\ if\ y\ =\ e^{Sin^{-1}\ x}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ W.\ K.\ T\ \frac{d}{dx}\ (e^x)\ =\ e^x\ and\ \frac{d}{dx}\ (Sin^{-1}\ x)\ =\ \frac{1}{\sqrt{1\ -\ x^2}}\ \hspace{10cm}\]
\[y\ =\ e^{Sin^{-1}\ x}\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ \frac{d}{dx}(e^{Sin^{-1}\ x})\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ e^{Sin^{-1}\ x}\ \frac{d}{dx}(Sin^{-1}\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ e^{Sin^{-1}\ x}\ \frac{1}{\sqrt{1\ -\ x^2}}\ \hspace{10cm}\]
\[Ex\ 11:\ Find\ \frac{dy}{dx}\ if\ y\ =\ log(\frac{1\ +\ Sin\ x}{1\ -\ Sin\ x})\ \hspace{15cm}\]
\[\color {black}{Solution:}\ W.\ K.\ T\ \frac{d}{dx}\ (log\ x)\ =\ \frac{1}{x}\ and\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[y\ =\ log(\frac{1\ +\ Sin\ x}{1\ -\ Sin\ x})\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ log(\frac{1\ +\ Sin\ x}{1\ -\ Sin\ x})\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{\frac{1\ +\ Sin\ x}{1\ -\ Sin\ x}}\ \frac{d}{dx}(\frac{1\ +\ Sin\ x}{1\ -\ Sin\ x})\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1\ -\ Sin\ x}{1\ +\ Sin\ x}\ [\frac{(1\ -\ Sin\ x)\ \frac{d}{dx}\ (1\ +\ Sin\ x)\ -\ (1\ +\ Sin\ x)\ \frac{d}{dx}\ (1\ -\ Sin\ x)}{(1\ -\ Sin\ x)^2}]\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1\ -\ Sin\ x}{1\ +\ Sin\ x}\ [\frac{(1\ -\ Sin\ x)\ (Cos\ x)\ -\ (1\ +\ Sin\ x)\ (-\ Cos\ x)}{(1\ -\ Sin\ x)^2}]\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1\ -\ Sin\ x}{1\ +\ Sin\ x}\ [\frac{Cos\ x\ -\ Cos\ x\ Sin\ x\ +\ Cos\ x\ +\ Cos\ x\ Sin\ x}{(1\ -\ Sin\ x)^2}]\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1\ -\ Sin\ x}{1\ +\ Sin\ x}\ [\frac{2\ Cos\ x}{(1\ -\ Sin\ x)^2}]\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{1\ +\ Sin\ x}\ [\frac{2\ Cos\ x}{(1\ -\ Sin\ x)}]\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{2\ Cos\ x}{1\ -\ Sin^2\ x}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{2\ Cos\ x}{Cos^2\ x}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Cos\ x}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ Sec\ x\ \hspace{10cm}\]
\[\color {royalblue} {Differentiation\ of\ Implicit\ functions}:\ \hspace{20cm}\]
\[\color {royalblue} {Examples}:\ \hspace{20cm}\]
\[Ex\ 1:\ Find\ \frac{dy}{dx}\ if\ x^2\ +\ y^2\ =\ 0\ \hspace{15cm}\]
\[\color {black}{Solution:}\ x^2\ +\ y^2\ =\ 0\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(x^2)\ +\ \frac{d}{dx}(y^2)\ =\ \frac{d}{dx}(0)\ \hspace{10cm}\]
\[2\ x\ +\ 2\ y\ \frac{d}{dx}(y)\ =\ 0\ \hspace{10cm}\]
\[2\ y\ \frac{dy}{dx}\ =\ -\ 2\ x\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{-\ x}{y}\ \hspace{10cm}\]
\[Ex\ 2:\ Find\ \frac{dy}{dx}\ if\ x^3\ +\ y^3\ =\ 3\ \hspace{15cm}\]
\[\color {black}{Solution:}\ x^3\ +\ y^3\ =\ 3\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(x^3)\ +\ \frac{d}{dx}(y^3)\ =\ \frac{d}{dx}(3)\ \hspace{10cm}\]
\[3\ x^2\ +\ 3\ y^2\ \frac{d}{dx}(y)\ =\ 0\ \hspace{10cm}\]
\[Divide\ by\ 3\ \hspace{10cm}\]
\[x^2\ +\ y^2\ \frac{dy}{dx}\ =\ 0\ \hspace{10cm}\]
\[y^2\ \frac{dy}{dx}\ =\ -\ x^2\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{-\ x^2}{y^2}\ \hspace{10cm}\]
\[Ex\ 3:\ Find\ \frac{dy}{dx}\ if\ x^2\ +\ y^2\ +\ 2\ x\ +\ 3\ y\ =\ 0\ \hspace{15cm}\]
\[\color {black}{Solution:}\ x^2\ +\ y^2\ +\ 2\ x\ +\ 3\ y\ =\ 0\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(x^2)\ +\ \frac{d}{dx}(y^2)\ +\ \frac{d}{dx}(2\ x)\ +\ \frac{d}{dx}(3\ y)\ =\ \frac{d}{dx}(0)\ \hspace{10cm}\]
\[2\ x\ +\ 2\ y\ \frac{d}{dx}(y)\ +\ 2\ (1)\ +\ 3\ \frac{d}{dx}(y)\ =\ 0\ \hspace{10cm}\]
\[\frac{dy}{dx}(2\ y\ +\ 3)\ =\ -\ 2\ x\ -\ 2\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ -\ \frac{2\ x\ +\ 2}{2\ y\ +\ 3}\ \hspace{10cm}\]
\[Ex\ 4:\ Find\ \frac{dy}{dx}\ if\ y\ =\ a\ +\ x\ e^y\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ a\ +\ x\ e^y\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(y)\ =\ \frac{d}{dx}(a\ +\ x\ e^y)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{d}{dx}(a)\ +\ \frac{d}{dx}(x\ e^y)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 0\ +\ x\ \frac{d}{dx}(e^y)\ +\ e^y\ \frac{d}{dx}(x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x\ e^y\ \frac{d}{dx}(y)\ +\ e^y\ (1)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x\ e^y\ \frac{dy}{dx}\ +\ e^y\ \hspace{10cm}\]
\[\frac{dy}{dx}(1\ -\ x\ e^y\ \frac{dy}{dx})\ =\ e^y\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ -\ \frac{e^y}{1\ -\ x\ e^y}\ \hspace{10cm}\]