# Examples On Differentiation Set-2

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Example - 29

If \begin{align}{x^2} + {y^2} = t + \frac{1}{t}\,\,{\rm{and}}\,\,{x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}\end{align}, then prove that \begin{align}\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{x^3}y}}\end{align}

Solution: We first try to use the two given relations to get rid of the parameter t , so that we obtain a (implicit) relation between x and y.

${x^2} + {y^2} = t + \frac{1}{t}$

Squaring, we get

${x^4} + {y^4} + 2{x^2}{y^2} = {t^2} + \frac{1}{{{t^2}}} + 2 \qquad\qquad\qquad... \text{(i)}$

Using the second relation in (i), we get

\begin{align}&2{x^2}{y^2} = 2\\ \Rightarrow \qquad & {y^2} = \frac{1}{{{x^2}}}\end{align}

Differentiating both sides w.r.t x, we get

\begin{align}&{\rm{ }}2y\frac{{dy}}{{dx}} = \frac{{ - 2}}{{{x^3}}}\\ \Rightarrow \qquad & \frac{{dy}}{{dx}} = \frac{{ - 1}}{{{x^3}y}}\end{align}

Example - 30

If $${y^2} = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x,$$ then prove that

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} + y = \frac{{{a^2}{b^2}}}{{{y^3}}}\end{align}

Solution: The final relation that we need to obtain is independent of sin x and cos x; this gives us a hint that using the given relation, we must first get rid of sin x and cos x:

\begin{align}{y^2} &= {a^2}{\cos ^2}x + {b^2}{\sin ^2}x\\& = \frac{1}{2}\left\{ {{a^2}\left( {2{{\cos }^2}x} \right) + {b^2}\left( {2{{\sin }^2}x} \right)} \right\}\\& = \frac{1}{2}\left\{ {{a^2}\left( {1 + \cos 2x} \right) + {b^2}\left( {1 - \cos 2x} \right)} \right\}\\ &= \frac{1}{2}\left\{ {\left( {{a^2} + {b^2}} \right) + \left( {{a^2} - {b^2}} \right)\cos 2x} \right\}\\& \Rightarrow 2{y^2} - \left( {{a^2} + {b^2}} \right) = \left( {{a^2} - {b^2}} \right)\cos 2x{\rm{ }}...{\rm{ }}\text{(i)} \end{align}

Differentiating both sides of (i) w.r.t x, we get

\begin{align}&{\rm{ }}4y\frac{{dy}}{{dx}} = - 2\left( {{a^2} - {b^2}} \right)\sin 2x\\ \Rightarrow \qquad &- 2y\frac{{dy}}{{dx}} = \left( {{a^2} - {b^2}} \right)\sin 2x{\rm{ }}\qquad...{\rm{ }} \text{(ii)}\end{align}

We see now that squaring (i) and (ii) and adding them will lead to an expression independent of the trig. terms:

${\left( {2{y^2} - \left( {{a^2} + {b^2}} \right)} \right)^2} + 4{y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} = {\left( {{a^2} - {b^2}} \right)^2}$

A slight rearrangement gives:

${\left( {\frac{{dy}}{{dx}}} \right)^2} + {y^2} - \left( {{a^2} + {b^2}} \right) = - \frac{{{a^2}{b^2}}}{{{y^2}}}{\rm{ }}\qquad...{\rm{ }}\text{(iii)}$

Differentiating both sides of (iii) w.r.t x:

\begin{align}&{\rm{ }}2\left( {\frac{{dy}}{{dx}}} \right)\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) + 2y\frac{{dy}}{{dx}} = \frac{{2{a^2}{b^2}}}{{{y^3}}}\frac{{dy}}{{dx}}\\ \Rightarrow \qquad &\frac{{{d^2}y}}{{d{x^2}}} + y = \frac{{{a^2}{b^2}}}{{{y^3}}}\end{align}

Example - 31

If the derivatives of f(x) and g(x) are known, find the derivative of $$y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}$$

Solution: We cannot directly differentiate the given relation since no rule tells us how to differentiate a term $${p^q}$$ where both p and q are variables.

What we can instead do is take the logarithm of both sides of the given relation:

\begin{align}&{\rm{ }}y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}\\ \Rightarrow \qquad &\ln y = g\left( x \right)\ln \left( {f\left( x \right)} \right) \end{align}

Now we differentiate both sides w.r.t x:

\begin{align}& \Rightarrow \frac{1}{y}\frac{{dy}}{{dx}} = g\left( x \right) \cdot \frac{1}{{f\left( x \right)}} \cdot f'\left( x \right) + \ln \left( {f\left( x \right)} \right) \cdot g'\left( x \right)\\& \Rightarrow \frac{{dy}}{{dx}} = y\left\{ {\frac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right.\\&{\rm{ }} = {\left( {f\left( x \right)} \right)^{g\left( x \right)}}\left\{ {\frac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right\}\end{align}

As a simple example, suppose we have to differentiate $$y={x^x}$$ :

\begin{align}&\quad\quad\quad\ln y = x\ln x\\& \Rightarrow \frac{1}{y}\frac{{dy}}{{dx}} = x \cdot \frac{1}{x} + \ln x \cdot 1\\&\qquad\qquad{\rm{ }} = 1 + \ln x\\&\quad \Rightarrow \frac{{dy}}{{dx}} = y\left( {1 + \ln x} \right)\\&\qquad\qquad{\rm{ }} = {x^x}\left( {1 + \ln x} \right)\end{align}