DIFFERENTIATION METHODS (Excercise Problems)

\[\LARGE{\color {purple} {PART- A}}\]
\[\color {purple} {1.}\ \color {red} {Find\ \frac{dy}{dx}}\ If\ y\ =\ log(Sec\ x)\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ y\ =\ log(Sec\ x)\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Sec\ x}\ \frac{d}{dx}(Sec\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Sin\ x}\ (Sec\ x\ Tan\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ Tan\ x\ \hspace{10cm}\]
\[\LARGE{\color {purple} {PART- B}}\]
\[\color {purple} {2.}\ If\ y\ =\ Sin^{-1}\ (x^2),\ \color {red} {find\ \frac{dy}{dx}}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ y\ =\ Sin^{-1}\ (x^2)\ \hspace{15cm}\]
\[W.\ K.\ T\ \frac{d}{dx}\ (Sin^{-1}\ x)\ =\ \frac{1}{\sqrt{1\ -\ x^2}}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{d}{dx}((Sin^{-1}\ (x^2))\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{\sqrt{1\ -\ (x^2)^2}}\ \frac{d}{dx}(x^2)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{\sqrt{1\ -\ x^4}}\ 2\ x\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{2\ x}{\sqrt{1\ -\ x^4}}\ \hspace{10cm}\]
\[\LARGE{\color {purple} {PART- C}}\]
\[\color {purple} {3:}\ \color {red} {Find\ \frac{dy}{dx}}\ If\ x^2\ siny\ =\ c\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ x^2\ siny\ =\ c\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(x^2\ siny)\ =\ \frac{d}{dx}(c)\ \hspace{10cm}\]
\[x^2\ \frac{d}{dx} (siny)\ +\ siny\frac{d}{dx} (x^2)\ =\ \ 0\ \hspace{10cm}\]
\[x^2\ cosy\ \frac{d}{dx} (y)\ +\ siny\ (2x)\ =\ \ 0\ \hspace{10cm}\]
\[x^2\ cosy\ \frac{dy}{dx}\ +\ 2x\ siny =\ 0\ \hspace{10cm}\]
\[x^2\ cosy\ \frac{dy}{dx}\ =\ \ -\ 2x\ siny\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{-\ 2x\ siny}{x^2\ cosy}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ -\ \frac{2}{x}\ tany\ \hspace{10cm}\]
\[\color {purple} {4:}\ \color {red} {Find\ \frac{dy}{dx}}\ If\ xy\ =\ k\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ xy\ =\ k\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(xy)\ =\ \frac{d}{dx}(k)\ \hspace{10cm}\]
\[x\ \frac{dy}{dx}\ +\ y(1)\ =\ \ 0\ \hspace{10cm}\]
\[x\ \frac{dy}{dx}\ =\ \ -\ y\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{-\ y}{x}\ \hspace{10cm}\]
\[\color {purple} {5:}\ \color {red} {Find\ \frac{dy}{dx}}\ if\ a\ x^2\ +\ 2h\ xy\ +\ b\ y^2\ +\ 2g\ x\ +\ 2f\ y\ +\ c\ =\ 0\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ a\ x^2\ +\ 2h\ xy\ +\ b\ y^2\ +\ 2g\ x\ +\ 2f\ y\ +\ c\ =\ 0\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[a\ \frac{d}{dx}(x^2)\ +\ 2h\ \frac{d}{dx}(xy)\ +\ b\ \frac{d}{dx}(y^2)\ +\ 2g\ \frac{d}{dx}(x)\ +\ 2f\ \frac{d}{dx}(y)\ +\ \frac{d}{dx}(c)\ =\ \frac{d}{dx}(0)\ \hspace{10cm}\]
\[2a\ x\ +\ 2h\ [x\ \frac{d}{dx}(y)\ +\ y\ \frac{d}{dx}(x)]\ +\ b\ 2y\ \frac{d}{dx}(y)\ +\ 2g\ (1)\ +\ 2f\ \frac{dy}{dx}\ +\ 0\ =\ 0\ \hspace{10cm}\]
\[2a\ x\ +\ 2h\ [x\ \frac{dy}{dx}\ +\ y\ (1)]\ +\ b\ 2y\ \frac{dy}{dx}\ +\ 2g\ +\ 2f\ \frac{dy}{dx}\ =\ 0\ \hspace{10cm}\]
\[\frac{dy}{dx}(2h\ x\ +\ 2b\ y\ +\ 2f)\ =\ -\ 2a\ x\ -\ 2h\ y\ -\ 2g\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ -\ \frac{ax\ +\ h\ y\ +\ g}{h\ x\ +\ b\ y\ +\ f}\ \hspace{10cm}\]
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