Question

If $y=e^x\sin x$ and $z=e^x\cos x$, then

• A. $y^{\text{'}\text{'}}=z$
• B. $y^{\text{'}\text{'}}=2z$
• C. $y^{\text{'}\text{'}}+z^{\text{'}\text{'}}=2(z-y)$
• D. $z^{\text{'}\text{'}}=-y$

Answer 1

The correct option is B

$y^{\text{'}\text{'}}=2z$

Explanation for the correct option.

Given that, $y=e^x\sin x$and $z=e^x\cos x$.

Differentiate $y$ using product rule i.e; $\frac{d\left(uv\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$.

$\Rightarrow y\text{'}=e^x\cos x+e^x\sin x$

Again differentiate $y\text{'}$ to get $y\text{'}\text{'}$.

$\begin{array}{rcl}{y}^{\text{'}\text{'}}& =& -e^x\sin x+e^x\cos x+e^x\cos x+e^x\sin x\\ & =& 2e^x\cos x\\ y\text{'}\text{'}& =& 2z\end{array}$

Hence, option B is correct.

Explanation for the incorrect options.

$z=e^x\cos x$

Differentiate $z$ using product rule i.e; $\frac{d\left(uv\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$.

$\Rightarrow z\text{'}=-e^x\sin x+e^x\cos x$

Again differentiate $z\text{'}$ to get $z\text{'}\text{'}$.

$\begin{array}{rcl}{z}^{\text{'}\text{'}}& =& -e^x\sin x+e^x\cos x-e^x\sin x-e^x\cos x\\ & =& -2e^x\sin x\\ z\text{'}\text{'}& =& -2y\end{array}$

Hence, all other options are incorrect.

Hence, B is the correct option.