DIFFERENTIATION METHODS (Excercise Problems)

\[\underline{PART\ -\ A}\]
\[1.\ Find\ \frac{dy}{dx}\ If\ y\ =\ log(Sec\ x)\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ log(Sec\ x)\ \hspace{15cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Sec\ x}\ \frac{d}{dx}(Sec\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{Sin\ x}\ (Sec\ x\ Tan\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ Tan\ x\ \hspace{10cm}\]
\[\underline{PART\ -\ B}\]
\[2.\ If\ y\ =\ Sin^{-1}\ (x^2),\ find\ \frac{dy}{dx}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ y\ =\ Sin^{-1}\ (x^2)\ \hspace{15cm}\]
\[W.\ K.\ T\ \frac{d}{dx}\ (Sin^{-1}\ x)\ =\ \frac{1}{\sqrt{1\ -\ x^2}}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{d}{dx}((Sin^{-1}\ (x^2))\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{\sqrt{1\ -\ (x^2)^2}}\ \frac{d}{dx}(x^2)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{\sqrt{1\ -\ x^4}}\ 2\ x\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{2\ x}{\sqrt{1\ -\ x^4}}\ \hspace{10cm}\]
\[\underline{PART\ -\ C}\]