Question

If $x = \cos \theta$ , $y = \sin 5\theta$ , then $\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}}$ is equal to
A. $- 5y$
B. $5y$
C. $25y$
D. $- 25y$

Answer 1

Hint : In order to find the value of $\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}}$ , split the equation into different parts, and simplify them each separately. Start with differentiating $x = \cos \theta$ and $y = \sin 5\theta$ with respect to $\theta$ , then divide the results in order to get $\dfrac{{dy}}{{dx}}$ . Substitute the values in the equation to solve and get the results.
Formula used:
$\dfrac{{d\left( {\cos \theta } \right)}}{{d\theta }} = - \sin \theta$
$\dfrac{{d\left( {\sin \theta } \right)}}{{d\theta }} = 5\cos 5\theta$
$\left( {1 - {{\cos }^2}\theta } \right) = {\sin ^2}\theta$
$\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta$
$\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$

Complete step-by-step answer :
We are given two functions: $x = \cos \theta$ and $y = \sin 5\theta$ .
Differentiating $x = \cos \theta$ with respect to $\theta$ , we get:
$\dfrac{{dx}}{{d\theta }} = \dfrac{{d\left( {\cos \theta } \right)}}{{d\theta }}$
$\Rightarrow \dfrac{{dx}}{{d\theta }} = \dfrac{{d\left( {\cos \theta } \right)}}{{d\theta }} = - \sin \theta$
$\Rightarrow \dfrac{{dx}}{{d\theta }} = - \sin \theta$ ……(1)
Similarly, differentiating $y = \sin 5\theta$ with respect to $\theta$ , we get:
$\dfrac{{dy}}{{d\theta }} = \dfrac{{d\left( {\sin 5\theta } \right)}}{{d\theta }}$
$\Rightarrow \dfrac{{dy}}{{d\theta }} = \dfrac{{d\left( {\sin \theta } \right)}}{{d\theta }} = 5\cos 5\theta$
$\Rightarrow \dfrac{{dy}}{{d\theta }} = 5\cos 5\theta$ …….(2)
Dividing equation 2 by equation 1, we get:
$\Rightarrow \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{5\cos 5\theta }}{{ - \sin \theta }}$
Cancelling $d\theta$ , we get:
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{5\cos 5\theta }}{{\sin \theta }}$ …..(3)
Since, we also need $\dfrac{{{d^2}y}}{{d{x^2}}}$ , so differentiating equation 3 again with respect to $x$ , we get:
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right)$
Substituting the value of equation 3 in above equation, we get:
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( { - \dfrac{{5\cos 5\theta }}{{\sin \theta }}} \right)$
Since, we have $\theta$ inside the brace, we can expand it as, where we divide and multiply by $d\theta$ , we get:
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{d\theta }}\left( { - \dfrac{{5\cos 5\theta }}{{\sin \theta }}} \right) \times \dfrac{{d\theta }}{{dx}}$
From the quotient rule, we know $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ .
Comparing $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right)$ with $\dfrac{d}{{d\theta }}\left( { - \dfrac{{5\cos 5\theta }}{{\sin \theta }}} \right)$ , we get:
$u = 5\cos 5\theta \\\ v = \sin \theta \;$
Expanding $\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{d\theta }}\left( { - \dfrac{{5\cos 5\theta }}{{\sin \theta }}} \right) \times \dfrac{{d\theta }}{{dx}}$ using $\dfrac{u}{v}$ method, we get:
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - 5\left( {\dfrac{{\sin \theta \dfrac{{d\left( {\cos 5\theta } \right)}}{{d\theta }} - \cos 5\theta \dfrac{{d\left( {\sin \theta } \right)}}{{d\theta }}}}{{{{\sin }^2}\theta }}} \right) \times \dfrac{1}{{ - \sin \theta }}$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - 5\left( {\dfrac{{\sin \theta \left( { - 5\sin 5\theta } \right) - \cos 5\theta \cos \theta }}{{{{\sin }^2}\theta }}} \right) \times \dfrac{1}{{ - \sin \theta }}$
Can be solved further and written as:
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \left( {\dfrac{{25\sin \theta \sin 5\theta + 5\cos 5\theta \cos \theta }}{{{{\sin }^3}\theta }}} \right)$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \left( {\dfrac{{ - 25\sin \theta \sin 5\theta - 5\cos 5\theta \cos \theta }}{{{{\sin }^3}\theta }}} \right)$
We need to find the value of $\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}}$ :
So, substituting the values we found in the above equation, we get:
$\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}}$
$\Rightarrow \left( {1 - {{\cos }^2}\theta } \right)\left( {\dfrac{{ - 25\sin \theta \sin 5\theta - 5\cos 5\theta \cos \theta }}{{{{\sin }^3}\theta }}} \right) - \cos \theta \left( { - \dfrac{{5\cos 5\theta }}{{\sin \theta }}} \right)$
Since, we know that $\left( {1 - {{\cos }^2}\theta } \right) = {\sin ^2}\theta$ , so writing that above:
$\Rightarrow \left( {{{\sin }^2}\theta } \right)\left( {\dfrac{{ - 25\sin \theta \sin 5\theta - 5\cos 5\theta \cos \theta }}{{{{\sin }^3}\theta }}} \right) + \dfrac{{5\cos \theta \cos 5\theta }}{{\sin \theta }}$
Cancelling ${\sin ^2}\theta$ :
$\Rightarrow \left( {\dfrac{{ - 25\sin \theta \sin 5\theta - 5\cos 5\theta \cos \theta }}{{\sin \theta }}} \right) + \dfrac{{5\cos \theta \cos 5\theta }}{{\sin \theta }}$
$\Rightarrow \dfrac{{ - 25\sin \theta \sin 5\theta }}{{\sin \theta }} - \dfrac{{5\cos 5\theta \cos \theta }}{{\sin \theta }} + \dfrac{{5\cos \theta \cos 5\theta }}{{\sin \theta }}$
Cancelling $\sin \theta$ from the first operand:
$\Rightarrow - 25\sin 5\theta - \dfrac{{5\cos 5\theta \cos \theta }}{{\sin \theta }} + \dfrac{{5\cos \theta \cos 5\theta }}{{\sin \theta }}$
Since, we know that $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta$ , so writing it in the above equation, we get:
$\Rightarrow - 25\sin 5\theta - 5\cos 5\theta \cot \theta + 5\cos 5\theta \cot \theta$
$\Rightarrow - 25\sin 5\theta$
As we were given that $y = \sin 5\theta$ , so writing $\sin 5\theta = y$ in the above equation, and we get:
$\Rightarrow - 25\sin 5\theta$
$\Rightarrow - 25y$
And, we get that $\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} = - 25y$ .
Therefore, Option 4 is correct.
So, the correct answer is “Option 4”.

Note : It’s important to remember the formulas of differentiation for easy solving and less error.
Always split the terms and differentiate or separately solve and get the result, solving at once may lead to error.