Random Sec 4 Differentiations

B6

y = 3e^x + \frac{4}{e^x}

\frac{dy}{dx} = 3e^x - \frac{4}{e^x}

\frac{d^2y}{dx^2} = 3e^x + \frac{4}{e^x}

let \frac{dy}{dx} = 0

3e^x - \frac{4}{e^x} = 0

3e^{2x} = 4

2x = \mathrm{ln} \frac{4}{3}

x = \frac{1}{2} \mathrm{ln} \frac{4}{3}

Sub x = \frac{1}{2} \mathrm{ln} \frac{4}{3} to \frac{d^2y}{dx^2}

\frac{d^2y}{dx^2} > 0 Thus, it is a min point.

C7

y = \mathrm{ln} \frac{5-4x}{3+2x}

y = \mathrm{ln} (5-4x) - \mathrm{ln} (3+2x)

\frac{dy}{dx} = \frac{-4}{5-4x} - \frac{2}{3+2x}

let \frac{dy}{dx} = 0

\frac{-4}{5-4x} - \frac{2}{3+2x} = 0

\frac{-4}{5-4x} = \frac{2}{3+2x}

-4(3+2x) = 2(5-4x)

-12 - 8x = 10 - 8x

-12 = 10 (NA).

There are no stationary points for this curve.

C8

x = \frac{1}{3}e^{y(2x+5)}

\mathrm{ln}(3x) = y(2x+5)

\frac{\mathrm{ln}(3x)}{2x+5} = y

y = \frac{\mathrm{ln}(3x)}{2x+5}

\frac{dy}{dx} = \frac{\frac{1}{x}(2x+5) - \mathrm{ln}(3x) \times 2}{(2x+5)^2}

Let x = e^2

\frac{dy}{dx} = \frac{\frac{1}{e^2}(2e^2+5) - \mathrm{ln}(3e^2) \times 2}{(2e^2+5)^2}

Evaluate with a calculator…

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