Question

If ${{\sin }^{2}}x+{{\cos }^{2}}y=1$, then $\dfrac{dy}{dx}$ is equal to
(a). $\dfrac{\sin 2x}{\sin 2y}$
(b). $\dfrac{{{\sin }^{2}}y}{\sin 2x}$
(c). $\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}y}$
(d). $-\dfrac{{{\sin }^{2}}y}{{{\sin }^{2}}x}$

Answer 1

Hint: Differentiate the given equation ${{\sin }^{2}}x+{{\cos }^{2}}y=1$ with respect to ‘x’. And take care of chain rule while differentiating ${{\sin }^{2}}x$ and ${{\cos }^{2}}y$ to get the solution accurately. Chain Rule is given as:
$\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right)g'\left( x \right)$

Complete step-by-step solution -

Given equation from the question is ${{\sin }^{2}}x+{{\cos }^{2}}y=1$ and we need to determine value of $'\dfrac{dy}{dx}'$ with the help of this equation. So, we need to differentiate the whole equation (both sides) with respect to x.
Hence, let us differentiate the whole equation given in the problem.
So, we get;
\[\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)+\dfrac{d}{dx}\left( {{\cos }^{2}}y \right)=\dfrac{d}{dx}\left( 1 \right).................\left( i \right)\]
Now, we know the property of differentiation given as;
$\dfrac{d}{dx}\left( {{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)+{{f}_{3}}\left( x \right)......{{f}_{n}}\left( x \right) \right)=\dfrac{d}{dx}\left( {{f}_{1}}\left( x \right) \right)+\dfrac{d}{dx}\left( {{f}_{2}}\left( x \right) \right)+\dfrac{d}{dx}\left( {{f}_{3}}\left( x \right) \right)......\dfrac{d}{dx}\left( {{f}_{n}}\left( x \right) \right)$
Hence, we can apply the above property with equation (i) and can re – write the equation as;
\[\dfrac{d}{dx}{{\left( \sin x \right)}^{2}}+\dfrac{d}{dx}{{\left( \cos x \right)}^{2}}=\dfrac{d}{dx}\left( 1 \right).................\left( ii \right)\]
As, we can observe that ${{\sin }^{2}}x$ and ${{\cos }^{2}}x$ are composite functions as ${{\sin }^{2}}x$ and ${{\cos }^{2}}x$ can also be written as ${{\left( \sin x \right)}^{2}}$ and ${{\left( \cos x \right)}^{2}}$. So, we can apply chain rule here to get differentiation.
Hence, now we need to use chain rule for differentiating ${{\sin }^{2}}x$ and ${{\cos }^{2}}x$. Chain rule is given as;
If we have a function of type f (g (x)), then derivative of the function can be given as;
$\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right).g'\left( x \right).............\left( iii \right)$
Hence, chain rule is applied to the composite functions and we need to differentiate it in continuous manner i.e. one by one.
So, we know the differentiation of ${{x}^{n}}$ and any constant term can be given as;
$\begin{align}
  & \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} \\
 & \dfrac{d}{dx}\left( \text{constant} \right)=0 \\
\end{align}$
Hence, equation (ii) can be simplified by applying the above relations. Hence, we get;
\[\begin{align}
  & 2{{\left( \sin x \right)}^{2-1}}\dfrac{d}{dx}\left( \sin x \right)+2{{\left( \cos y \right)}^{2-1}}\dfrac{d}{dx}\left( \cos y \right)=0 \\
 & or \\
 & 2\sin x\dfrac{d}{dx}\left( \sin x \right)+2\cos y\dfrac{d}{dx}\left( \cos y \right)=0............\left( iv \right) \\
\end{align}\]
Now, we know \[\dfrac{d}{dx}\left( \sin x \right)\ and\ \dfrac{d}{dx}\left( \cos x \right)\] can be given as $\cos x\ and\ -\sin x.$ Hence, equation (iv) can be written by putting values of \[\dfrac{d}{dx}\left( \sin x \right)\ and\ \dfrac{d}{dx}\left( \cos x \right)\]. Hence, we get;
$\begin{align}
  & 2\sin x\cos x+2\cos y\left( -\sin y \right)\dfrac{dy}{dx}=0 \\
 & 2\sin x\cos x-2\cos y\sin y\dfrac{dy}{dx}=0..................\left( v \right) \\
\end{align}$
Now, we can use trigonometric identity $\sin 2x=2\sin x\cos x$ in the above equation. Here, we can observe that $2\sin x\cos x$ can be replaced by $\sin 2x$ and $2\sin y\cos y$ can be replaced by $\sin 2y$.
Hence, we can further simplify the equation (v) as;
$\begin{align}
  & \sin 2x-\sin 2y\dfrac{dy}{dx}=0 \\
 & \dfrac{dy}{dx}=\dfrac{\sin 2x}{\sin 2y} \\
\end{align}$
Hence, if ${{\sin }^{2}}x+{{\cos }^{2}}y=1$, then $\dfrac{dy}{dx}$can be given as $\dfrac{dy}{dx}=\dfrac{\sin 2x}{\sin 2y}$.
So, option ‘A’ is the correct answer.

Note: One can go wrong with the differentiation of ${{\sin }^{2}}x$ and ${{\cos }^{2}}y$. He/she may miss the concept of chain rule and can write, $\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)={{\cos }^{2}}x\ \ or\ \ \dfrac{d}{dx}\left( {{\cos }^{2}}y \right)=-{{\sin }^{2}}y\dfrac{dy}{dx}$ which are wrong. So, take care of chain rule with the composite functions.
One can get confusion in the derivative of $\sin x$ and $\cos x$. He/she may write $\dfrac{d}{dx}\left( \sin x \right)=-\cos x\ \ and\ \ \dfrac{d}{dx}\left( \cos x \right)=\sin x$ which are wrong. Hence, take care of derivative formulae as well.