1. Differentiate $y = \sin^{-1}(3x)$. Step 1: Recall the derivative formula for $y = \sin^{-1}(u)$ is $\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$. Step 2: Here, $u = 3x$, so $\frac{du}{dx} = 3$. Step 3: Substitute into the formula: $$\frac{dy}{dx} = \frac{3}{\sqrt{1-(3x)^2}} = \frac{3}{\sqrt{1-9x^2}}.$$ 2. Find the gradient of $y = (x^2 + 5) \ln(\sin(x^3 + 8x + 7))$. Step 1: Use the product rule: $\frac{dy}{dx} = u'v + uv'$, where $u = x^2 + 5$ and $v = \ln(\sin(x^3 + 8x + 7))$. Step 2: Compute $u' = 2x$. Step 3: Compute $v' = \frac{1}{\sin(x^3 + 8x + 7)} \cdot \cos(x^3 + 8x + 7) \cdot (3x^2 + 8)$ by chain rule. Step 4: Substitute: $$\frac{dy}{dx} = 2x \ln(\sin(x^3 + 8x + 7)) + (x^2 + 5) \cdot \frac{\cos(x^3 + 8x + 7)(3x^2 + 8)}{\sin(x^3 + 8x + 7)}.$$ 3. If $y = \log_{10} x$, find $\frac{dy}{dx}$. Step 1: Recall $\log_{10} x = \frac{\ln x}{\ln 10}$. Step 2: Differentiate: $$\frac{dy}{dx} = \frac{1}{x \ln 10}.$$ 4. Find $\frac{dy}{dx}$ of $y = (x-1)(x-2)(x-3)(x-4)$. Step 1: Use logarithmic differentiation or product rule repeatedly. Step 2: Alternatively, expand partially or use sum of derivatives: $$\frac{dy}{dx} = \sum_{i=1}^4 \prod_{j \neq i} (x - j).$$ Step 3: Explicitly, $$\frac{dy}{dx} = (x-2)(x-3)(x-4) + (x-1)(x-3)(x-4) + (x-1)(x-2)(x-4) + (x-1)(x-2)(x-3).$$ 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) $y = \sqrt[3]{4t} = (4t)^{1/3}$ $$\frac{dy}{dt} = \frac{1}{3} (4t)^{-2/3} \times 4 = \frac{4}{3} (4t)^{-2/3}.$$ 6. Find derivatives: (a) $y = (23 + x)^5$ $$\frac{dy}{dx} = 5 (23 + x)^4.$$ (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) $y = (1 - w)^{-1}$ $$\frac{dy}{dw} = -1 \times (1 - w)^{-2} \times (-1) = (1 - w)^{-2}.$$ 7. Given $y = 4^{4x} + 6^x$, find $\frac{dy}{dx}$. Step 1: Use derivative of $a^{u} = a^{u} \ln a \frac{du}{dx}$. Step 2: For $4^{4x}$, $u = 4x$, $\frac{du}{dx} = 4$. Step 3: Differentiate: $$\frac{dy}{dx} = 4^{4x} \ln 4 \times 4 + 6^x \ln 6 = 4 \times 4^{4x} \ln 4 + 6^x \ln 6.$$ 8. Given $y = \sum_{r=0}^{10} (r!) x^{r+1}$, find $\frac{dy}{dx}$. Step 1: Differentiate term-by-term: $$\frac{dy}{dx} = \sum_{r=0}^{10} (r!) (r+1) x^r.$$ 9. Find $\frac{dy}{dx}$ if $x^3 - 4x^3 y + 3xy + y + 5xy^2 = 17$. Step 1: Differentiate both sides implicitly: $$3x^2 - 4(3x^2 y + x^3 \frac{dy}{dx}) + 3(y + x \frac{dy}{dx}) + \frac{dy}{dx} + 5(y^2 + 2x y \frac{dy}{dx}) = 0.$$ Step 2: Collect terms with $\frac{dy}{dx}$ and solve for $\frac{dy}{dx}$. 10. If $y = x^{2x}$, find $\frac{dy}{dx}$. Step 1: Rewrite $y = e^{2x \ln x}$. Step 2: Differentiate: $$\frac{dy}{dx} = y \cdot \frac{d}{dx}(2x \ln x) = x^{2x} (2 \ln x + 2).$$ 11. Given $y = \sin x$, find constants $a$ and $b$ if $$x \frac{d^3 y}{dx^3} + b \frac{dy}{dx} + a y = 0.$$ Step 1: Compute derivatives: $$\frac{dy}{dx} = \cos x, \quad \frac{d^2 y}{dx^2} = -\sin x, \quad \frac{d^3 y}{dx^3} = -\cos x.$$ Step 2: Substitute: $$x (-\cos x) + b \cos x + a \sin x = 0.$$ Step 3: For this to hold for all $x$, coefficients must satisfy: $$-x \cos x + b \cos x + a \sin x = 0 \implies (b - x) \cos x + a \sin x = 0.$$ Step 4: This cannot hold for all $x$ unless $a = 0$ and $b = x$ (not constant). So no constant $a,b$ satisfy this. 12. Find gradient of $y = \ln(\tan 3x)$. Step 1: Use chain rule: $$\frac{dy}{dx} = \frac{1}{\tan 3x} \cdot \sec^2 3x \cdot 3 = 3 \frac{\sec^2 3x}{\tan 3x}.$$