Re: Calculus and Exponents

I'd write it as $\displaystyle \displaystyle \begin{align*} y = e^3 \left[ \sin{(x)} + 1 \right] ^{-1} \end{align*}$ then let $\displaystyle \displaystyle \begin{align*} u = \sin{(x)} + 1 \implies y = e^3 \, u^{-1} \end{align*}$. Then $\displaystyle \displaystyle \begin{align*} \frac{du}{dx} = \cos{(x)} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \frac{dy}{du} = -e^3 \, u^{-2} = -e^3 \left[ \sin{(x)} + 1 \right] ^{-2} \end{align*}$. So $\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = -e^3 \cos{(x)} \left[ \sin{(x)} + 1 \right] ^{-2} = -\frac{ e^3 \cos{(x)} }{ \left[ \sin{(x)} + 1 \right] ^2} \end{align*}$. Your answer is correct.

Re: Calculus and Exponents

Your friend failed to realize that $\displaystyle e^3$ is a constant, and so its derivative is zero.