Question

If $x = {\cos ^7}\theta$and $y = \sin \theta$, then $\dfrac{{{d^3}x}}{{d{y^3}}} =$
(A) $\dfrac{{105}}{4}\sin 4\theta$
(B) $\dfrac{{105}}{2}\sin 2\theta$
(C) $\dfrac{{105}}{4}\cos 4\theta$
(D) None of these

Answer 1

In this question, we have to choose the required solution for the given particular options.
First, we have to differentiate the given x and y terms with respect to x. by using the chain rule we can differentiate the given terms. The given of the question is the value of x and y, they are the functions of trigonometry terms. Here we have to use some trigonometric and differentiation formula to find the required solution. We have mentioned that in the formula used. Then using the all required terms in the required solution, we will get the correct result.

Complete step-by-step answer:
It is given that, $x = {\cos ^7}\theta$and $y = \sin \theta$.
We need to find out the value of $\dfrac{{{d^3}x}}{{d{y^3}}}$.
Now,$x = {\cos ^7}\theta$
Differentiating x with respect to $\theta$, we get,
$\dfrac{{dx}}{{d\theta }} = 7{\cos ^6}\theta \left( { - \sin \theta } \right) = - 7\sin \theta {\cos ^6}\theta$……….i)
$y = \sin \theta$
Differentiating y with respect to $\theta$, we get,
\Rightarrow$$$\dfrac{{dy}}{{d\theta }} = \cos \theta $$…………ii) Here, y depends on the variable \theta $and also the variable x depends on the variable$\theta $, then applying chain rule we get,$\Rightarrow\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d\theta }} \times \dfrac{{d\theta }}{{dx}}$$ Using i) and ii) we get, $\Rightarrow\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d\theta }} \times \dfrac{{\dfrac{1}{1}}}{{\dfrac{{dx}}{{d\theta }}}} = \cos \theta \times \dfrac{1}{{ - 7\sin \theta {{\cos }^6}\theta }} = \dfrac{1}{{ - 7\sin \theta {{\cos }^5}\theta }}$v$\dfrac{{dx}}{{dy}} = - 7\sin \theta {\cos ^5}\theta Differentiating with respect to y, we get, $\Rightarrow$$$\dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{d}{{dy}}\left( { - 7\sin \theta {{\cos }^5}\theta } \right)
Let us consider that $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d\theta }} \times \dfrac{{d\theta }}{{dx}}$ we get,
$= \dfrac{{d\theta }}{{dy}}.\dfrac{d}{{d\theta }}\left( { - 7\sin \theta {{\cos }^5}\theta } \right)$
Using ii) ,we get,
$= \dfrac{1}{{\cos \theta }}\left( { - 7\cos \theta {{\cos }^5}\theta - 7\sin \theta \times 5{{\cos }^4}\theta \left( { - \sin \theta } \right)} \right)$
if$y = uv,\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}}v + u\dfrac{{dv}}{{dx}}$ then,
$= - 7{\cos ^5}\theta + 35{\sin ^2}\theta {\cos ^3}\theta$
Since we already know that ${\sin ^2}\theta = 1 - {\cos ^2}\theta$
$= - 7{\cos ^5}\theta + 35\left( {1 - {{\cos }^2}\theta } \right){\cos ^3}\theta$
Solving we get,
$= - 7{\cos ^5}\theta + 35{\cos ^3}\theta - 35{\cos ^5}\theta$
Thus we get, $\dfrac{{{d^2}x}}{{d{y^2}}} = - 42{\cos ^5}\theta + 35{\cos ^3}\theta$
Differentiating $\dfrac{{{d^2}x}}{{d{y^2}}}$ with respect to y we get,
$\dfrac{{{d^3}x}}{{d{y^3}}} = \dfrac{d}{{dy}}\left( { - 42{{\cos }^5}\theta + 35{{\cos }^3}\theta } \right)$
Let us consider that $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d\theta }} \times \dfrac{{d\theta }}{{dx}}$ we get,
$\dfrac{{{d^3}x}}{{d{y^3}}} = \dfrac{{d\theta }}{{dy}} \times \dfrac{d}{{d\theta }}\left( { - 42{{\cos }^5}\theta + 35{{\cos }^3}\theta } \right)$
Using ii) we get,
$= \dfrac{1}{{\cos \theta }}\left( { - 42 \times 5{{\cos }^4}\theta \left( { - \sin \theta } \right) + 35 \times 3{{\cos }^2}\theta \left( { - \sin \theta } \right)} \right)$
$= 210\sin \theta {\cos ^3}\theta - 105\sin \theta \cos \theta$
Using the trigonometric formula, $\cos 2\theta = 2{\cos ^2}\theta - 1$
$= 105\sin \theta \cos \theta \left( {2{{\cos }^2}\theta - 1} \right)$
Using the trigonometric formula, $\sin 2\theta = 2\sin \theta \cos \theta$
$= \dfrac{{105}}{2} \times 2\sin \theta \cos \theta \times \cos 2\theta$
Simplifying we get,
$= \dfrac{{105}}{2}\sin 2\theta \cos 2\theta$
Multiply and divide by 2 we get,
$= \dfrac{{105}}{4} \times 2\sin 2\theta \cos 2\theta$
Hence we get,
$= \dfrac{{105}}{4}\sin 4\theta$

$\therefore$ The option (A) is the correct option.

Note: Chain Rule:
If a variable y depends on the variable z, which itself depends on the variable x (that is y and z are dependent variables), then y, via the intermediate variable of z, depends on x as well. In which case, the chain rule states that: $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dz}} \times \dfrac{{dz}}{{dx}}$.
Differentiation formula:
$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
$\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$
Trigonometric formula:
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\cos 2\theta = 2{\cos ^2}\theta - 1$
$\sin 2\theta = 2\sin \theta \cos \theta$