Question Statement

Find $\frac{dy}{dx}$ given:

$y=\frac{e^{2x}}{e^{-2x}+1}$

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Background and Explanation

This problem involves differentiating a quotient of exponential functions. You will need to apply the quotient rule along with the chain rule for differentiating composite exponential functions of the form $e^{ax}$.

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Solution

We are given the function:

$y=\frac{e^{2x}}{e^{-2x}+1}$

To differentiate this with respect to $x$, we identify the numerator and denominator:

• Let $u=e^{2x}$ (so $\frac{du}{dx}=2e^{2x}$ by the chain rule)
• Let $v=e^{-2x}+1$ (so $\frac{dv}{dx}=-2e^{-2x}$ by the chain rule)

Applying the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{(e^{-2x}+1)\cdot \frac{d}{dx}(e^{2x})-e^{2x}\cdot \frac{d}{dx}(e^{-2x}+1)}{(e^{-2x}+1{)}^2}\\ & =\frac{(e^{-2x}+1)\cdot e^{2x}\cdot 2-e^{2x}\cdot (e^{-2x}(-2)+0)}{(e^{-2x}+1{)}^2}\\ & =\frac{2e^{-2x}\cdot e^{2x}+2e^{2x}+2e^{2x}\cdot e^{-2x}}{(e^{-2x}+1{)}^2}\\ & =\frac{2e^{-2x+2x}+2e^{2x}+2e^{2x-2x}}{(e^{-2x}+1{)}^2}\\ & =\frac{2e^0+2e^{2x}+2e^0}{(e^{-2x}+1{)}^2}\\ & =\frac{2+2e^{2x}+2}{(e^{-2x}+1{)}^2}\\ \frac{dy}{dx}& =\frac{4+2e^{2x}}{(e^{-2x}+1{)}^2}\end{array}$

Explanation of key steps:

• In the second line, we compute the derivatives: $\frac{d}{dx}(e^{2x})=2e^{2x}$ and $\frac{d}{dx}(e^{-2x}+1)=-2e^{-2x}$
• In the third line, we distribute $(e^{-2x}+1)\cdot 2e^{2x}=2e^{-2x}{e}^{2x}+2e^{2x}$ and simplify $-e^{2x}\cdot (-2e^{-2x})=+2e^{2x}{e}^{-2x}$
• Lines 4-5 use the exponent law $e^a\cdot e^b=e^{a+b}$ and the fact that $e^0=1$
• The final line combines the constant terms $2+2=4$

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Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
• Chain Rule for Exponentials: $\frac{d}{dx}(e^{ax})=a{e}^{ax}$
• Exponent Laws: $e^a\cdot e^b=e^{a+b}$ and $e^0=1$

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Summary of Steps

1. Identify components: Let $u=e^{2x}$ and $v=e^{-2x}+1$
2. Differentiate: Compute $\frac{du}{dx}=2e^{2x}$ and $\frac{dv}{dx}=-2e^{-2x}$ using the chain rule
3. Apply quotient rule: Substitute into $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$
4. Expand numerator: Distribute $(e^{-2x}+1)\cdot 2e^{2x}$ to get $2e^{-2x}{e}^{2x}+2e^{2x}$ and simplify $-e^{2x}\cdot (-2e^{-2x})=+2e^{2x}{e}^{-2x}$
5. Simplify exponents: Use $e^{-2x}\cdot e^{2x}=e^0=1$ to reduce the expression to $\frac{2+2e^{2x}+2}{(e^{-2x}+1{)}^2}$
6. Combine constants: Add $2+2=4$ to obtain the final result $\frac{4+2e^{2x}}{(e^{-2x}+1{)}^2}$