1. Differentiate $y = \sin^{-1}(3x)$. Using the chain rule, the derivative of $\sin^{-1}(u)$ is $\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$. Here, $u = 3x$, so $\frac{du}{dx} = 3$. Therefore, $$\frac{dy}{dx} = \frac{3}{\sqrt{1-(3x)^2}} = \frac{3}{\sqrt{1-9x^2}}.$$ 2. Find the gradient of $y = (x^2 + 5) \log_a \sin(x^3 + 8x + 7)$. Use the product rule: $\frac{d}{dx}[uv] = u'v + uv'$. Let $u = x^2 + 5$, so $u' = 2x$. Let $v = \log_a \sin(x^3 + 8x + 7)$. Recall $\log_a f = \frac{\ln f}{\ln a}$, so $$v' = \frac{1}{\ln a} \cdot \frac{1}{\sin(x^3 + 8x + 7)} \cdot \cos(x^3 + 8x + 7) \cdot (3x^2 + 8)$$ by chain rule. Therefore, $$\frac{dy}{dx} = 2x \cdot \log_a \sin(x^3 + 8x + 7) + (x^2 + 5) \cdot \frac{\cos(x^3 + 8x + 7)(3x^2 + 8)}{\sin(x^3 + 8x + 7) \ln a}.$$ 3. If $y = \log_{10} x$, find $\frac{dy}{dx}$. Recall $\frac{d}{dx} \log_a x = \frac{1}{x \ln a}$. So, $$\frac{dy}{dx} = \frac{1}{x \ln 10}.$$ 4. Find $\frac{dy}{dx}$ of $y = (x - 1)(x - 2)(x - 3)(x - 4)$. Use product rule or expand first. Expanding: $$(x-1)(x-2) = x^2 - 3x + 2,$$ $$(x-3)(x-4) = x^2 - 7x + 12,$$ So, $$y = (x^2 - 3x + 2)(x^2 - 7x + 12).$$ Multiply: $$y = x^4 - 7x^3 + 12x^2 - 3x^3 + 21x^2 - 36x + 2x^2 - 14x + 24,$$ Simplify: $$y = x^4 - 10x^3 + 35x^2 - 50x + 24.$$ Differentiate termwise: $$\frac{dy}{dx} = 4x^3 - 30x^2 + 70x - 50.$$ 5. Find derivatives: (a) $y = 5w^{3.5}$ $$\frac{dy}{dw} = 5 \times 3.5 w^{2.5} = 17.5 w^{2.5}.$$ (b) $y = \frac{2.4}{p^{3.4}} = 2.4 p^{-3.4}$ $$\frac{dy}{dp} = 2.4 \times (-3.4) p^{-4.4} = -8.16 p^{-4.4}.$$ (c) $y = \frac{1}{\sqrt{2 s^5}} = (2 s^5)^{-1/2} = 2^{-1/2} s^{-5/2}$ $$\frac{dy}{ds} = 2^{-1/2} \times (-\frac{5}{2}) s^{-7/2} = -\frac{5}{2 \sqrt{2}} s^{-7/2}.$$ (d) $y = \sqrt[3]{4t} = (4t)^{1/3}$ $$\frac{dy}{dt} = \frac{1}{3} (4t)^{-2/3} \times 4 = \frac{4}{3} (4t)^{-2/3} = \frac{4}{3 \cdot 4^{2/3} t^{2/3}} = \frac{4^{1/3}}{3 t^{2/3}}.$$ 6. Find derivatives: (a) $y = (23x + 1)^5$ $$\frac{dy}{dx} = 5 (23x + 1)^4 \times 23 = 115 (23x + 1)^4.$$ (b) $y = \frac{1}{(52 - t)^2} = (52 - t)^{-2}$ $$\frac{dy}{dt} = -2 (52 - t)^{-3} \times (-1) = 2 (52 - t)^{-3}.$$ (c) $y = \sqrt[3]{21 + a} = (21 + a)^{1/3}$ $$\frac{dy}{da} = \frac{1}{3} (21 + a)^{-2/3}.$$ (d) $y = \sin 3\phi$ $$\frac{dy}{d\phi} = 3 \cos 3\phi.$$ (e) $y = \cos(32) \phi = (\cos 32) \phi$ (assuming $\cos 32$ is constant) $$\frac{dy}{d\phi} = \cos 32.$$ (f) $y = (1 - w)^{-1}$ $$\frac{dy}{dw} = -1 (1 - w)^{-2} \times (-1) = (1 - w)^{-2}.$$ 7. Given $y = 4^{4x} + 6^x$, find $\frac{dy}{dx}$. Recall $\frac{d}{dx} a^{u} = a^{u} \ln a \frac{du}{dx}$. For $4^{4x}$, $u = 4x$, $\frac{du}{dx} = 4$. So, $$\frac{d}{dx} 4^{4x} = 4^{4x} \ln 4 \times 4 = 4 \cdot 4^{4x} \ln 4.$$ For $6^x$, $$\frac{d}{dx} 6^x = 6^x \ln 6.$$ Therefore, $$\frac{dy}{dx} = 4 \cdot 4^{4x} \ln 4 + 6^x \ln 6.$$ 8. Given $y = \sum_{r=0}^{10} (r!) x^{r+1}$, find $\frac{dy}{dx}$. Differentiate termwise: $$\frac{dy}{dx} = \sum_{r=0}^{10} (r!) (r+1) x^r.$$ 9. Find $\frac{dy}{dx}$ if $$x^3 - 4x^3 y + 3xy + y + 5xy^2 = 17.$$ Implicit differentiation: Differentiate both sides w.r.t $x$: $$3x^2 - 4(3x^2 y + x^3 \frac{dy}{dx}) + 3(y + x \frac{dy}{dx}) + \frac{dy}{dx} + 5(y^2 + 2x y \frac{dy}{dx}) = 0,$$ Simplify: $$3x^2 - 12x^2 y - 4x^3 \frac{dy}{dx} + 3y + 3x \frac{dy}{dx} + \frac{dy}{dx} + 5y^2 + 10x y \frac{dy}{dx} = 0.$$ Group $\frac{dy}{dx}$ terms: $$(-4x^3 + 3x + 1 + 10x y) \frac{dy}{dx} = -3x^2 + 12x^2 y - 3y - 5y^2.$$ Therefore, $$\frac{dy}{dx} = \frac{-3x^2 + 12x^2 y - 3y - 5y^2}{-4x^3 + 3x + 1 + 10x y}.$$ 10. If $y = x^{2x}$, find $\frac{dy}{dx}$. Rewrite: $$y = e^{\ln(x^{2x})} = e^{2x \ln x}.$$ Differentiate: $$\frac{dy}{dx} = y \cdot \frac{d}{dx} (2x \ln x) = x^{2x} \cdot (2 \ln x + 2) = 2 x^{2x} (\ln x + 1).$$ 11. Given $y = \sin x$, find constants $a$ and $b$ such that $$x \frac{d^3 y}{dx^3} + b \frac{dy}{dx} + a y = 0.$$ Calculate derivatives: $$\frac{dy}{dx} = \cos x,$$ $$\frac{d^2 y}{dx^2} = -\sin x,$$ $$\frac{d^3 y}{dx^3} = -\cos x.$$ Substitute: $$x (-\cos x) + b \cos x + a \sin x = 0,$$ or $$-x \cos x + b \cos x + a \sin x = 0.$$ For this to hold for all $x$, coefficients of $\sin x$ and $\cos x$ must be zero separately: Coefficient of $\sin x$: $a = 0$. Coefficient of $\cos x$: $-x + b = 0$ for all $x$ is impossible unless $b$ depends on $x$. Since $b$ is constant, no solution unless $b = 0$ and $x \cos x = 0$ for all $x$, which is false. Hence, no constants $a,b$ satisfy this for all $x$ except trivial $a=0,b=0$. 12. Find gradient of $y = \log_a (\tan 3x)$. Recall $\frac{d}{dx} \log_a u = \frac{1}{u \ln a} \frac{du}{dx}$. Here, $u = \tan 3x$, so $$\frac{du}{dx} = 3 \sec^2 3x.$$ Therefore, $$\frac{dy}{dx} = \frac{1}{\tan 3x \ln a} \times 3 \sec^2 3x = \frac{3 \sec^2 3x}{\tan 3x \ln a}.$$