Chapter 5: Continuity and Differentiability

Ans:  The given function is  $ \text{y=}\left| \begin{align}  & \text{f(x)}\,\,\,\text{g(x)}\,\,\,\text{h(x)} \\  & \text{   l}\ \ \ \ \ \ \text{m    n} \\  & \text{   a       b    c  } \\ \end{align} \right| $ 

Evaluate the determinant.

 $ \text{y=(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x)} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

 $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (mc-nb)f(x) }\!\!]\!\!\text{ -}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (lc-na)g(x) }\!\!]\!\!\text{ +}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (lb-ma)h(x) }\!\!]\!\!\text{ } \\ & \text{=(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x)} \\  & =\left| \begin{matrix}\text{f}'\left(\text{x}\right)&\text{g}'\left(\text{x}\right)&\text{h}'\left(\text{x} \right)  \\\text{l} & \text{m} & \text{n}  \\\text{a} & \text{b} & \text{c}  \\\end{matrix} \right| \\ \end{align} $ 

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\left|\begin{matrix}\text{f}'\left(\text{x}\right)&\text{g }'\left( \text{x} \right) & \text{h }'\left( \text{x} \right)  \\\text{l} & \text{m} & \text{n}  \\\text{a} & \text{b} & \text{c}  \\\end{matrix} \right| $

Ans: The given equation is

$\text{y=}{{\text{e}}^{\text{aco}{{\text{s}}^{\text{-1}}}\text{x}}} $ .

Then take the logarithm on both sides of the equation.

$\begin{align}&\text{logy=aco}{{\text{s}}^{\text{-1}}}\text{xloge}\\&\Rightarrow\text{logy=aco}{{\text{s}}^{\text{-1}}}\text{x} \\ \end{align} $ 

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=ax}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\\&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-ax}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \\ \end{align} $ 

Therefore, squaring both sides of the equation gives

$\begin{align}&{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}\dfrac{{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}}{\text{1-}{{\text{x}}^{\text{2}}}}\\&\Rightarrow(\text{1-}{{\text{x}}^{\text{2}}}\text{)}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}\\&\Rightarrow\text{(1-}{{\text{x}}^{\text{2}}}\text{)}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}} \\ \end{align} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives  $\begin{align}&{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\dfrac{\text{d}}{\text{dx}}\text{(1-}{{\text{x}}^{\text{2}}}\text{)+(1-}{{\text{x}}^{\text{2}}}\text{)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\left[{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\right]\text{=}{{\text{a}}^{\text{2}}}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{y}}^{\text{2}}}\text{)}\\&\Rightarrow{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{(-2x)+(1-}{{\text{x}}^{\text{2}}}\text{)}\!\!\times\!\!\text{2}\dfrac{\text{dy}}{\text{dx}}\text{}\!\!\times\!\!\text{}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\text{a}}^{\text{2}}}\text{}\!\!\times\!\!\text{2y}\!\!\times\!\!\text{}\dfrac{\text{dy}}{\text{dx}}\\&\Rightarrow\text{x}\dfrac{\text{dy}}{\text{dx}}\text{+(1-}{{\text{x}}^{\text{2}}}\text{)}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\text{a}}^{\text{2}}}\text{}\!\!\times\!\!\text{ y} \\ \end{align} $

Hence, it is proved that  $\text{(1-}{{\text{x}}^{\text{2}}}\text{)}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-x}\dfrac{\text{dy}}{\text{dx}}\text{-}{{\text{a}}^{\text{2}}}\text{y=0} $