Question

If \[y = a\sin x + b\cos x\], then \[{y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2}\] is a
(a)Function of \[x\]
(b)Function of \[y\]
(c)Function of \[x\] and \[y\]
(d)None of these

Answer 1

Hint: We will find the value of \[\dfrac{{dy}}{{dx}}\] and square it. We will also find the value of \[{y^2}\]. We will find the sum of these two values. We will simplify the equation and check whether the expression obtained is a function of \[x\], \[y\], both or none. We will choose the correct option accordingly.

Complete step by step solution:
First, we will find the differentiation of function\[y\]:
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {a\sin x + b\cos x} \right)\]
We know that \[\dfrac{d}{{dx}}\left( {c + d} \right) = \dfrac{d}{{dx}}\left( c \right) + \dfrac{d}{{dx}}\left( d \right)\].
We will substitute \[a\sin x\] for \[c\] and \[b\cos x\] for \[d\] in the above formula.
   \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {a\sin x} \right) + \dfrac{d}{{dx}}\left( {b\cos x} \right)\].
We know that \[\dfrac{d}{{dx}}\left( {px} \right) = p\dfrac{d}{{dx}}\left( x \right)\] where \[p\] is constant.
We will substitute \[a\] and \[b\] for \[p\] in the above formula:
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = a\dfrac{d}{{dx}}\left( {\sin x} \right) + b\dfrac{d}{{dx}}\left( {\cos x} \right)\]
We know that \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\] and \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]. We will use these results in the above equation:
 \[\begin{array}{l} \Rightarrow \dfrac{{dy}}{{dx}} = a\cos x + b\left( { - \sin x} \right)\\ \Rightarrow \dfrac{{dy}}{{dx}} = a\cos x - b\sin x\end{array}\]
We will find the square of \[\dfrac{{dy}}{{dx}}\]:
\[{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {\left( {a\cos x - b\sin x} \right)^2}\]
We know that \[{\left( {p - q} \right)^2} = {p^2} - 2pq + {q^2}\]. We will substitute \[a\cos x\] for \[p\] and \[b\sin x\] for \[q\] in the formula for the square of the difference between 2 numbers:
 \[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {\left( {a\cos x} \right)^2} - 2\left( {a\cos x} \right)\left( {b\sin x} \right) + {\left( {b\sin x} \right)^2}\]
We will simplify this equation:
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {a^2}{\cos ^2}x - 2ab\sin x\cos x + {b^2}{\sin ^2}x\].
We will find the value of \[{y^2}\]:
 \[{y^2} = {\left( {a\sin x + b\cos x} \right)^2}\]
We know that \[{\left( {p + q} \right)^2} = {p^2} + 2pq + {q^2}\]. We will substitute \[a\sin x\] for \[p\] and \[b\cos x\] for \[q\] in the formula for the square of the sum of 2 numbers:
 \[ \Rightarrow {y^2} = {\left( {a\sin x} \right)^2} + 2\left( {a\sin x} \right)\left( {b\cos x} \right) + {\left( {b\cos x} \right)^2}\]
We will simplify this equation:
\[ \Rightarrow {y^2} = {a^2}{\sin ^2}x + 2ab\sin x\cos x + {b^2}{\cos ^2}x\]
We will find the sum of \[{y^2}\] and \[{\left( {\dfrac{{dy}}{{dx}}} \right)^2}\]:
 \[ \Rightarrow {y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {a^2}{\sin ^2}x + 2ab\sin x\cos x + {b^2}{\cos ^2}x + {a^2}{\cos ^2}x - 2ab\sin x\cos x + {b^2}{\sin ^2}x\]
We will collect the like terms:
\[ \Rightarrow {y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \left( {{a^2}{{\sin }^2}x + {b^2}{{\sin }^2}x} \right) + \left( {2ab\sin x\cos x - 2ab\sin x\cos x} \right) + \left( {{b^2}{{\cos }^2}x + {a^2}{{\cos }^2}x} \right)\]
We will take out the common factor and simplify the equation:
\[\begin{array}{l} \Rightarrow {y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \left( {{a^2} + {b^2}} \right){\sin ^2}x + 0 + \left( {{b^2} + {a^2}} \right){\cos ^2}x\\ \Rightarrow {y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \left( {{a^2} + {b^2}} \right){\sin ^2}x + \left( {{a^2} + {b^2}} \right){\cos ^2}x\\ \Rightarrow {y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \left( {{a^2} + {b^2}} \right)\left( {{{\sin }^2} + {{\cos }^2}x} \right)\end{array}\]
We will substitute 1 for \[\left( {{{\sin }^2} + {{\cos }^2}x} \right)\] in the above equation:
\[\begin{array}{l} \Rightarrow {y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \left( {{a^2} + {b^2}} \right) \cdot 1\\ \Rightarrow {y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {a^2} + {b^2}\end{array}\]
We can see that \[{y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2}\] is not a function \[x\], \[y\] or both \[x\] and \[y\].
$\therefore $ Option (d) is the correct option.

Note: We should be careful of the fact that \[{\left( {\dfrac{{dy}}{{dx}}} \right)^2}\] is not the same as \[\dfrac{{{d^2}y}}{{d{y^2}}}\]. \[{\left( {\dfrac{{dy}}{{dx}}} \right)^2}\]is the square of the differentiation of the function \[y\] whereas \[\dfrac{{{d^2}y}}{{d{y^2}}}\] is the second differentiation of the function \[y\]. \[\dfrac{{dy}}{{dx}}\] is also written as \[y'\] whereas \[\dfrac{{{d^2}y}}{{d{y^2}}}\] is written as \[y''\]. Also we need to know the basic properties of differentiation so that we can solve the question easily.