Apply the power rule (d)/(dx)(axⁿ) = anxⁿ−1 and the constant rule (d)/(dx)(c) = 0 to each term.

$$\frac{dy}{dx} = \frac{d}{dx}(3x^4) - \frac{d}{dx}(5x^2) + \frac{d}{dx}(7x) - \frac{d}{dx}(2)$$ $$\frac{dy}{dx} = 3(4)x^{4-1} - 5(2)x^{2-1} + 7(1)x^{1-1} - 0$$

$$\frac{dy}{dx} = 12x^3 - 10x^1 + 7x^0$$ $$\frac{dy}{dx} = 12x^3 - 10x + 7$$

The correct option is (A).
$\boxed{\text{A}}$
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2. If $y = (2x + 3)^5$, find $dy/dx$.

$$\frac{dy}{dx} = 5(2x + 3)^4 \cdot 2$$ $$\frac{dy}{dx} = 10(2x + 3)^4$$

The correct option is (B).
$\boxed{\text{B}}$
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3. Differentiate $y = x^3 \sin(x)$ with respect to $x$.

First, find $u'$ and $v'$.

$$u = x^3 \implies u' = \frac{d}{dx}(x^3) = 3x^2$$ $$v = \sin(x) \implies v' = \frac{d}{dx}(\sin(x)) = \cos(x)$$

$$\frac{dy}{dx} = (3x^2)(\sin(x)) + (x^3)(\cos(x))$$ $$\frac{dy}{dx} = 3x^2 \sin(x) + x^3 \cos(x)$$

The correct option is (D).
$\boxed{\text{D}}$
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4. Find the derivative of $y = (3x - 1) / (2x + 5)$.

First, find $u'$ and $v'$.

$$u = 3x - 1 \implies u' = \frac{d}{dx}(3x - 1) = 3$$ $$v = 2x + 5 \implies v' = \frac{d}{dx}(2x + 5) = 2$$

$$\frac{dy}{dx} = \frac{6x + 15 - (6x - 2)}{(2x + 5)^2}$$ $$\frac{dy}{dx} = \frac{6x + 15 - 6x + 2}{(2x + 5)^2}$$ $$\frac{dy}{dx} = \frac{17}{(2x + 5)^2}$$

The correct option is (A).
$\boxed{\text{A}}$
Drop the next question.

5. Evaluate $dy/dx$ at $x = 2$ for the curve $y = 2x^3 - 4x + 5$.

$$\frac{dy}{dx} = \frac{d}{dx}(2x^3) - \frac{d}{dx}(4x) + \frac{d}{dx}(5)$$ $$\frac{dy}{dx} = 2(3)x^{3-1} - 4(1)x^{1-1} + 0$$ $$\frac{dy}{dx} = 6x^2 - 4$$

$$\frac{dy}{dx}\Big|_{x=2} = 6(2)^2 - 4$$ $$\frac{dy}{dx}\Big|_{x=2} = 6(4) - 4$$ $$\frac{dy}{dx}\Big|_{x=2} = 24 - 4$$ $$\frac{dy}{dx}\Big|_{x=2} = 20$$

The correct option is (A).
$\boxed{\text{A}}$
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6. Find the gradient of the curve $y = x^2 - 3x + 2$ at the point where $x = 4$.

$$\frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(2)$$ $$\frac{dy}{dx} = 2x^{2-1} - 3(1)x^{1-1} + 0$$ $$\frac{dy}{dx} = 2x - 3$$

$$\text{Gradient} = 2(4) - 3$$ $$\text{Gradient} = 8 - 3$$ $$\text{Gradient} = 5$$

The correct option is (A).
$\boxed{\text{A}}$
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7. Find the stationary points (turning points) of the curve $y = x^3 - 3x + 4$.

$$\frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x) + \frac{d}{dx}(4)$$ $$\frac{dy}{dx} = 3x^2 - 3$$

$$3x^2 - 3 = 0$$ $$3x^2 = 3$$ $$x^2 = 1$$ $$x = \pm\sqrt{1}$$ $$x = 1 \quad \text{or} \quad x = -1$$

The correct option is (A).
$\boxed{\text{A}}$
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8. Determine the minimum value of the function $y = x^2 - 6x + 10$.

$$\frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(6x) + \frac{d}{dx}(10)$$ $$\frac{dy}{dx} = 2x - 6$$

$$2x - 6 = 0$$ $$2x = 6$$ $$x = 3$$

$$y = (3)^2 - 6(3) + 10$$ $$y = 9 - 18 + 10$$ $$y = -9 + 10$$ $$y = 1$$

The correct option is (A).
$\boxed{\text{A}}$
Drop the next question.

9. If the displacement of a particle is given by $s = 2t^3 - 9t^2 + 12t + 5$, find its acceleration when $t = 2$ seconds.

$$v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t^2 + 12t + 5)$$ $$v = 2(3)t^2 - 9(2)t^1 + 12(1)t^0 + 0$$ $$v = 6t^2 - 18t + 12$$

$$a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 18t + 12)$$ $$a = 6(2)t^1 - 18(1)t^0 + 0$$ $$a = 12t - 18$$

$$a = 12(2) - 18$$ $$a = 24 - 18$$ $$a = 6 \text{ m/s}^2$$

The correct option is (A).
$\boxed{\text{A}}$
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10. Differentiate $y = \cos(4x^3 - 2x)$ with respect to $x$.

First, find $\frac{du}{dx}$.

$$u = 4x^3 - 2x \implies \frac{du}{dx} = \frac{d}{dx}(4x^3 - 2x) = 4(3)x^2 - 2(1) = 12x^2 - 2$$

$$\frac{dy}{dx} = -\sin(4x^3 - 2x) \cdot (12x^2 - 2)$$ $$\frac{dy}{dx} = -(12x^2 - 2) \sin(4x^3 - 2x)$$

The correct option is (A).
$\boxed{\text{A}}$
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11. Find $dy/dx$ if $y = e^{5x^2}$.

First, find $\frac{du}{dx}$.

$$u = 5x^2 \implies \frac{du}{dx} = \frac{d}{dx}(5x^2) = 5(2)x^{2-1} = 10x$$

$$\frac{dy}{dx} = e^{5x^2} \cdot (10x)$$ $$\frac{dy}{dx} = 10x e^{5x^2}$$

The correct option is (A).
$\boxed{\text{A}}$
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12. Differentiate $y = \ln(3x^2 + 4)$ with respect to $x$.

First, find $\frac{du}{dx}$.

$$u = 3x^2 + 4 \implies \frac{du}{dx} = \frac{d}{dx}(3x^2 + 4) = 3(2)x^{2-1} + 0 = 6x$$

$$\frac{dy}{dx} = \frac{1}{3x^2 + 4} \cdot (6x)$$ $$\frac{dy}{dx} = \frac{6x}{3x^2 + 4}$$

The correct option is (A).
$\boxed{\text{A}}$
Drop the next question.

13. Given the implicit equation $x^2 + y^2 = 25$, find $dy/dx$.

Remember that $y$ is a function of $x$, so use the chain rule for $y^2$.

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$ $$2x + 2y \frac{dy}{dx} = 0$$

$$2y \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = \frac{-2x}{2y}$$ $$\frac{dy}{dx} = -\frac{x}{y}$$

The correct option is (A).
$\boxed{\text{A}}$
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14. Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 7$ cm.

$$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2)$$ $$\frac{dA}{dr} = \pi (2r)$$ $$\frac{dA}{dr} = 2\pi r$$

$$\frac{dA}{dr}\Big|_{r=7} = 2\pi (7)$$ $$\frac{dA}{dr}\Big|_{r=7} = 14\pi \text{ cm}^2/\text{cm}$$

The correct option is (A).
$\boxed{\text{A}}$
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15. If $y = x \tan(x)$, find $dy/dx$.

First, find $u'$ and $v'$.

$$u = x \implies u' = \frac{d}{dx}(x) = 1$$ $$v = \tan(x) \implies v' = \frac{d}{dx}(\tan(x)) = \sec^2(x)$$

$$\frac{dy}{dx} = (1)(\tan(x)) + (x)(\sec^2(x))$$ $$\frac{dy}{dx} = \tan(x) + x \sec^2(x)$$

The correct option is (A).
$\boxed{\text{A}}$