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 xand \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}}}

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)
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