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