Tutorial 6 (rough)

Rough work for MTH1008 Personal Tutoring on 2023-10-30

$$y=\ln \left(\frac{\cosh x-1}{\cosh x+1}\right)=\ln (\cosh x-1)-\ln (\cosh x+1)$$

$$⟹\frac{dy}{dx}=\frac{\sinh x}{\cosh x-1}-\frac{\sinh x}{\cosh x+1}$$

$$⟹\frac{dy}{dx}=\frac{\sinh x(\cosh x+1)-\sinh x(\cosh x-1)}{{\sinh }^{2}x}$$

$$⟹\frac{dy}{dx}=\frac{2\sinh x}{{\sinh }^{2}x}$$

$$⟹\frac{dy}{dx}=2\csc x$$

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$$1+\frac{3}{4x-8}-\frac{3}{4x+8}⟹1+\frac{3(4x+8)-3(4x-8)}{16x^2-64}$$

$$⟹1+\frac{48}{16x^2-64}⟹\frac{16x^2-16}{16x^2-64}⟹\frac{x^2-1}{x^2-4}$$

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$$y=\ln (x^2\sqrt{1-x^2})⟹y=2\ln x+\frac{1}{2}\ln (1-x^2)$$

$$\frac{dy}{dx}=\frac{2}{x}-\frac{x}{1-x^2}⟹\frac{dy}{dx}=\frac{2(1-x^2)-x^2}{x(1-x^2)}$$

$$⟹\frac{dy}{dx}=\frac{2-3x^2}{x(1-x^2)}$$

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$$y=\frac{e^{2x}\cdot \ln x}{(x-3{)}^3}$$

$$⟹\ln y=\ln (e^{2x}\cdot \ln x)-3\ln (x-3)$$

$$⟹\ln y=2x+\ln (\ln x)-3\ln (x-3)$$

$$⟹\frac{1}{y}\cdot \frac{dy}{dx}=2+\frac{1}{x\ln x}-\frac{3}{x-3}$$

$$\frac{dy}{dx}=\left(\frac{e^{2x}\cdot \ln x}{(x-3{)}^3}\right)\cdot \left(2+\frac{1}{x\ln x}-\frac{3}{x-3}\right)$$

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Find an expression for the slope on the lower-half of the circle:

$$x^2+y^2=25⟹2x+2y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{x}{y}:y=-\sqrt{25-x^2}⟹\frac{x}{\sqrt{25-x^2}}$$

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Find the slope of a general point of the expression

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

$$⟹\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$$

$$⟹\frac{dy}{dx}=-\frac{x\cdot b^2}{y\cdot a^2}$$

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