Calculus and Exponents

• May 2nd 2013, 01:27 AM
lukasaurus
Calculus and Exponents
A question says to find the derivative of

$e^3/(sin(x)+1)$

The answer I got was $-e^3cosx/(sinx+1)^2$

My friend got
$(e^3sinx+e^3 - e^3cosx)/(sinx+1)^2$

The online calculators all agree with me, but he used the quotient rule (so did I), but I pulled e^3 out of it and used 1/sinx+1 to perform the quotient rule.

Which is the correct method?
• May 2nd 2013, 01:45 AM
Prove It
Re: Calculus and Exponents
I'd write it as \displaystyle \begin{align*} y = e^3 \left[ \sin{(x)} + 1 \right] ^{-1} \end{align*} then let \displaystyle \begin{align*} u = \sin{(x)} + 1 \implies y = e^3 \, u^{-1} \end{align*}. Then \displaystyle \begin{align*} \frac{du}{dx} = \cos{(x)} \end{align*} and \displaystyle \begin{align*} \frac{dy}{du} = -e^3 \, u^{-2} = -e^3 \left[ \sin{(x)} + 1 \right] ^{-2} \end{align*}. So \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.
• May 2nd 2013, 01:55 AM
MarkFL
Re: Calculus and Exponents
Your friend failed to realize that $e^3$ is a constant, and so its derivative is zero.