You're on a roll — here are the solutions to your questions: 1. Solve $\frac{dy}{dx}$ if $y = \frac{e^{8x}}{1+e^{4x}}$ when $y=1$ and $x=0$. Step 1: Use the quotient rule for differentiation, $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$, where $u = e^{8x}$ and $v = 1+e^{4x}$. First, find the derivatives of $u$ and $v$: $$ \frac{du}{dx} = \frac{d}{dx}(e^{8x}) = 8e^{8x} $$ $$ \frac{dv}{dx} = \frac{d}{dx}(1+e^{4x}) = 4e^{4x} $$ Step 2: Apply the quotient rule. $$ \frac{dy}{dx} = \frac{(1+e^{4x})(8e^{8x}) - (e^{8x})(4e^{4x})}{(1+e^{4x})^2} $$ Step 3: Simplify the expression. $$ \frac{dy}{dx} = \frac{8e^{8x} + 8e^{12x} - 4e^{12x}}{(1+e^{4x})^2} $$ $$ \frac{dy}{dx} = \frac{8e^{8x} + 4e^{12x}}{(1+e^{4x})^2} $$ Step 4: Evaluate $\frac{dy}{dx}$ at $x=0$. Substitute $x=0$ into the derivative: $$ \frac{dy}{dx}\Big|_{x=0} = \frac{8e^{8(0)} + 4e^{12(0)}}{(1+e^{4(0)})^2} $$ $$ \frac{dy}{dx}\Big|_{x=0} = \frac{8e^0 + 4e^0}{(1+e^0)^2} $$ Since $e^0 = 1$: $$ \frac{dy}{dx}\Big|_{x=0} = \frac{8(1) + 4(1)}{(1+1)^2} = \frac{8+4}{2^2} = \frac{12}{4} = 3 $$ The condition $y=1$ at $x=0$ is not consistent with the given function $y(0) = \frac{e^0}{1+e^0} = \frac{1}{2}$. We proceed by evaluating the derivative at $x=0$. The final answer is $\boxed{3}$. 2. Solve $\frac{dy}{dx}$ if $y = \ln\left(\frac{x+1}{x-1}\right)$. Step 1: Use the logarithm property $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$. $$ y = \ln(x+1) - \ln(x-1) $$ Step 2: Differentiate each term with respect to $x$. Recall that $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$. $$ \frac{dy}{dx} = \frac{d}{dx}(\ln(x+1)) - \frac{d}{dx}(\ln(x-1)) $$ $$ \frac{dy}{dx} = \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) - \frac{1}{x-1} \cdot \frac{d}{dx}(x-1) $$ $$ \frac{dy}{dx} = \frac{1}{x+1}(1) - \frac{1}{x-1}(1) $$ $$ \frac{dy}{dx} = \frac{1}{x+1} - \frac{1}{x-1} $$ Step 3: Combine the fractions by finding a common denominator. $$ \frac{dy}{dx} = \frac{(x-1) - (x+1)}{(x+1)(x-1)} $$ $$ \frac{dy}{dx} = \frac{x-1-x-1}{x^2-1} $$ $$ \frac{dy}{dx} = \frac{-2}{x^2-1} $$ The final answer is $\boxed{\frac{-2}{x^2-1}}$. 3. Find intercepts and asymptotes for $h(x) = \frac{2x-4}{x^3-1}$. a) x-intercepts: Set $h(x) = 0$. This occurs when the numerator is zero. $$ 2x-4 = 0 $$ $$ 2x = 4 $$ $$ x = 2 $$ The x-intercept is $\boxed{(2, 0)}$. b) y-intercepts: Set $x = 0$. $$ h(0) = \frac{2(0)-4}{0^3-1} = \frac{-4}{-1} = 4 $$ The y-intercept is $\boxed{(0, 4)}$. c) Vertical asymptotes: Set the denominator to zero. $$ x^3-1 = 0 $$ Factor the difference of cubes: $(x-1)(x^2+x+1) = 0$. The real root is $x-1=0 \implies x=1$. The quadratic factor $x^2+x+1$ has a discriminant $\Delta = 1^2 - 4(1)(1) = -3 < 0$, so it has no other real roots. Check that the numerator is not zero at $x=1$: $2(1)-4 = -2 \ne 0$. The vertical asymptote is $\boxed{x=1}$. d) Horizontal asymptotes: Compare the degrees of the numerator and denominator. Degree of numerator ($N$) = 1. Degree of denominator ($D$) = 3. Since $N < D$, the horizontal asymptote is $y=0$. The horizontal asymptote is $\boxed{y=0}$. e) Slant asymptotes: A slant asymptote exists if the degree of the numerator is exactly one more than the degree of the denominator ($N = D+1$). In this case, $N=1$ and $D=3$, so $N \ne D+1$. There is $\boxed{\text{no slant asymptote}}$. Got