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Question: If \[y = a\sin x + b\cos x\], then \[{y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2}\] is a A.Func...

If y=asin⁡x+bcos⁡xy = a\sin x + b\cos x, then y2+(dydx)2{y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} is a
A.Function of xx
B.Function of yy
C.Function of xx and yy
D.Constant

Explanation

Solution

Hint : The process of finding a derivative is called differentiation. The reverse process is called antidifferentiation. The fundamental theorem of calculus relates antidifferentiation with integration. The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument.

Complete step-by-step answer :
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The tangent line is the best linear approximation of the function near that input value. For this reason, the derivative is often described as the "instantaneous rate of change", the ratio of the instantaneous change in the dependent variable to that of the independent variable.
ddx(sin⁡x)=cos⁡x\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x
ddx(cos⁡x)=−sin⁡x\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x
Here in this question we have y=asin⁡x+bcos⁡xy = a\sin x + b\cos x
Differentiating both the sides with respect to xx we get ,
dydx=acos⁡x−bsin⁡x\dfrac{{dy}}{{dx}} = a\cos x - b\sin x
Therefore y2+(dydx)2=(asin⁡x+bcos⁡x)2+(acos⁡x−bsin⁡x)2{y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {\left( {a\sin x + b\cos x} \right)^2} + {\left( {a\cos x - b\sin x} \right)^2}
Simplifying the equation we get ,
y2+(dydx)2=a2sin⁡2x+b2cos⁡2x+2absin⁡xcos⁡x+a2cos⁡2x+b2sin⁡2x−2absin⁡xcos⁡x{y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {a^2}{\sin ^2}x + {b^2}{\cos ^2}x + 2ab\sin x\cos x + {a^2}{\cos ^2}x + {b^2}{\sin ^2}x - 2ab\sin x\cos x
Hence on further simplification we get ,
y2+(dydx)2=a2(sin⁡2x+cos⁡2x)+b2(cos⁡2x+sin⁡2x){y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {a^2}\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {b^2}\left( {{{\cos }^2}x + {{\sin }^2}x} \right)
Since sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
Hence we get
y2+(dydx)2=a2+b2{y^2} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {a^2} + {b^2} (which is a constant)
Therefore option (4) is the correct answer.
So, the correct answer is “Option 4”.

Note : The process of finding a derivative is called differentiation. The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument.