## Calculus: Early Transcendentals (2nd Edition)

$\frac{dy}{dx} = \frac{2^x\ln(2)}{(2^x+1)^2}$
If $f(x) = b^x$, then $f'(x) = b^x\ln(b)$ $y = \frac{2^x}{2^x+1}$ Chain Rule and Quotient Rule: $\frac{dy}{dx} = \frac{(2^x\ln(2))(2^x+1)-(2^x)(2^x\ln(2))}{(2^x+1)^2} = \frac{2^x\ln(2)(2^x+1-2^x)}{(2^x+1)^2} = \frac{2^x\ln(2)}{(2^x+1)^2}$