Question

If $y=a\sin x+bc\mathrm{os}x$, then $y^2+{\left(\frac{dy}{dx}\right)}^2$ is equal to?

• A. function of $x$
• B. function of $y$
• C. function of $x$ and $y$
• D. constant

Answer 1

The correct option is D

constant

Explanation for the correct option:

Step 1. Find the value of $\frac{dy}{dx}$.

Differentiate the function $y=a\sin x+bc\mathrm{os}x$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}(a\sin x+b\cos x)\\ & =& a\cos x+b(-\sin x)\mathbf{}\left[\because \frac{d}{\mathrm{dx}}\left(\mathrm{sinx}\right)=\mathrm{cosx},\frac{d}{\mathrm{dx}}\left(\mathrm{cosx}\right)=-\mathrm{sinx}\right]\\ & =& a\cos x-b\sin x\end{array}$

Step 2. Find the value of the expression.

The expression $y^2+{\left(\frac{dy}{dx}\right)}^2$ can be evaluated as:

$\begin{array}{rcl}{y}^2+{\left(\frac{dy}{dx}\right)}^2& =& {\left(a\sin x+b\cos x\right)}^2+{\left(a\cos x-b\sin x\right)}^2\\ & =& a^2{\sin }^{2}x+2ab\sin x\cos x+b^2{\cos }^{2}x+a^2{\cos }^{2}x-2ab\sin x\cos x+b^2{\sin }^{2}x\\ & =& a^2\left({\sin }^{2}x+{\cos }^{2}x\right)+b^2\left({\sin }^{2}x+{\cos }^{2}x\right)\\ & =& a^2+b^2\end{array}$

And as $a$ and $b$ are constants, so the value of $y^2+{\left(\frac{dy}{dx}\right)}^2$ is also a constant.

Hence, the correct option is D.