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Section 2.3, #26: Compute the derivative of $f(x) = \dfrac{x^2+5x+6}{x^2-4}$.
Solution: We use the quotient rule to see $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{x^2+5x+6}{x^2-4} \right] &= \dfrac{(x^2-4) \frac{\mathrm{d}}{\mathrm{d}x}[x^2+5x+6] - (x^2+5x+6) \frac{\mathrm{d}}{\mathrm{d}x}[x^2-4]}{(x^2-4)^2} \\ &=\dfrac{(x^2-4)(2x+5) - (x^2+5x+6)(2x)}{(x^2-4)^2} \\ &=\dfrac{2x^3+5x^2-8x-20-2x^3-10x^2-12x}{(x^2-4)^2} \\ &=\dfrac{-5x^2-20x-20}{(x^2-4)^2} \\ &=\dfrac{-5(x^2+4x+4)}{(x-2)^2(x+2)^2} \\ &=\dfrac{-5(x+2)^2}{(x-2)^2(x+2)^2} \\ &=-\dfrac{5}{(x-2)^2}. \end{array}$$

Section 2.3, #98: Find the second derivative of $f(x)=\sec(x)$.
Solution: We know the derivative of $\sec(x)$ is $$\dfrac{\mathrm{d}}{\mathrm{d}x} \sec(x)=\sec(x)\tan(x)$$ and we know that the derivative of $\tan(x)$ is $$\dfrac{\mathrm{d}}{\mathrm{d}x} \tan(x) = \sec^2(x).$$ Now we compute $$\begin{array}{ll} \dfrac{\mathrm{d}^2}{\mathrm{d}x^2} \sec(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \sec(x)\tan(x) \right] \\ &= \sec'(x)\tan(x) + \sec(x)\tan'(x) \\ &= \sec(x)\tan^2(x)+\sec^3(x). \end{array}$$

Section 2.4, #8: Find the derivative of $y=5(2-x^3)^4$.
Solution: First compute the first derivative using the chain rule to get $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left[5\left(2-x^3\right)^4 \right] = 20 \left(2-x^3 \right)^3(-3x^2)=-60x^2(2-x^3)^3.$$

Section 2.4, #27: Find the derivative of $y= \dfrac{x}{\sqrt{x^2+1}}$.
Solution: First rewrite this and replace the square root with an exponent: $$y=\dfrac{x}{(x^2+1)^{\frac{1}{2}}}.$$ Now proceed with the quotient rule and the chain rule to get $$\begin{array}{ll} \dfrac{\mathrm{d}y}{\mathrm{d}x} &= \dfrac{(x^2+1)^{\frac{1}{2}} \cdot 1 - x \left(\frac{1}{2}\right)(x^2+1)^{-\frac{1}{2}}(2x+0)}{[(x^2+1)^{\frac{1}{2}})^2} \\ &= \dfrac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{(x^2+1)} \\ &= \dfrac{(x^2+1)-x^2}{(x^2+1)^{\frac{3}{2}}} \\ &=\dfrac{1}{(x^2+1)^{\frac{3}{2}}}. \end{array}$$

Section 2.4, #63: Find the derivative of $y=\sin(\tan(2x))$.
Solution: Using the chain rule and the rules for differentiating sine and tangent, we get $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \sin(\tan(2x)) &= \cos(\tan(2x)) \dfrac{\mathrm{d}}{\mathrm{d}x} \tan(2x) \\ &=\cos(\tan(2x)) \sec^2(2x) \dfrac{\mathrm{d}}{\mathrm{d}x} [2x] \\ &= 2 \cos(\tan(2x)) \sec^2(2x). \end{array}$$