ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 6- Worked Out Examples

Saturday, 2 August 2014

CHAPTER 6- Worked Out Examples

Example: 1

If ${x^2} + {y^2} = t + \dfrac{1}{t}\,\,$ and $\,{x^4} + {y^4} = {t^2} + \dfrac{1}{{{t^2}}}$ , then prove that $\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^3}y}}$

Solution: 1

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 $y$ .

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

Squaring, we get

${x^4} + {y^4} + 2{x^2}{y^2} = {t^2} + \dfrac{1}{{{t^2}}} + 2$

$\ldots(i)$

Using the second relation in ( $i$ ), we get

	$2{x^2}{y^2} = 2$
	$\Rightarrow \,\, {y^2} = \dfrac{1}{{{x^2}}}$

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

	$2y\dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{{x^3}}}$
	$\Rightarrow \,\, \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^3}y}}$

Example: 2

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

$\dfrac{{{d^2}y}}{{d{x^2}}} + y = \dfrac{{{a^2}{b^2}}}{{{y^3}}}$

Solution: 2

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 $\cos x$ :

	${y^2} = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x$
	$= \dfrac{1}{2}\left\{ {{a^2}\left( {2{{\cos }^2}x} \right) + {b^2}\left( {2{{\sin }^2}x} \right)} \right\}$
	$= \dfrac{1}{2}\left\{ {{a^2}\left( {1 + \cos 2x} \right) + {b^2}\left( {1 - \cos 2x} \right)} \right\}$
	$= \dfrac{1}{2}\left\{ {\left( {{a^2} + {b^2}} \right) + \left( {{a^2} - {b^2}} \right)\cos 2x} \right\}$
	$\Rightarrow \, 2{y^2} - \left( {{a^2} + {b^2}} \right) = \left( {{a^2} - {b^2}} \right)\cos 2x$	$\ldots(i)$

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

	$4y\dfrac{{dy}}{{dx}} = - 2\left( {{a^2} - {b^2}} \right)\sin 2x$
	$\Rightarrow \,\, - 2y\dfrac{{dy}}{{dx}} = \left( {{a^2} - {b^2}} \right)\sin 2x$	$\ldots(ii)$

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

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

A slight rearrangement gives:

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

$\ldots(iii)$

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

	$2\left( {\dfrac{{dy}}{{dx}}} \right)\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) + 2y\dfrac{{dy}}{{dx}} = \dfrac{{2{a^2}{b^2}}}{{{y^3}}}\dfrac{{dy}}{{dx}}$
	$\Rightarrow \,\, \dfrac{{{d^2}y}}{{d{x^2}}} + y = \dfrac{{{a^2}{b^2}}}{{{y^3}}}$

Example:3

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: 3

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:

	$y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}$
	$\Rightarrow \,\, \ln y = g\left( x \right)\ln \left( {f\left( x \right)} \right)$

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

	$\Rightarrow \,\, \dfrac{1}{y}\dfrac{{dy}}{{dx}} = g\left( x \right) \cdot \dfrac{1}{{f\left( x \right)}} \cdot f'\left( x \right) + \ln \left( {f\left( x \right)} \right) \cdot g'\left( x \right)$
	$\Rightarrow \,\, \dfrac{{dy}}{{dx}} = y\left\{ {\dfrac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right.$
	$= {\left( {f\left( x \right)} \right)^{g\left( x \right)}}\left\{ {\dfrac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right\}$

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

	$\ln y = x\ln x$
	$\Rightarrow \,\, \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \cdot \dfrac{1}{x} + \ln x \cdot 1$
	$= 1 + \ln x$
	$\Rightarrow \,\, \dfrac{{dy}}{{dx}} = y\left( {1 + \ln x} \right)$
	$= {x^x}\left( {1 + \ln x} \right)$

Saturday, 2 August 2014

CHAPTER 6- Worked Out Examples

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