Practice Questions: Rules for Differentiation

$$\begin{align*} \frac{dy}{dx} &= 3 \cdot \frac{d}{dx}[x^2] \\ &= 3 \cdot 2x \\ &= 6x \end{align*}$$

$$\begin{align*} f'(x) &= 25 \cdot \frac{d}{dx}[x] \\ &= 25 \cdot 1 \\ &= 25 \end{align*}$$

The derivative of any constant is zero.

$$\begin{align*} k'(x) &= 0 \end{align*}$$

$$\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}[-4x^5] + \frac{d}{dx}[2] \\ &= -4 \cdot 5x^4 + 0 \\ &= -20x^4 \end{align*}$$

$$\begin{align*} g'(x) &= 16 \cdot (-2)x^{-2-1} \\ &= -32x^{-3} \\ &= -\frac{32}{x^3} \end{align*}$$

$$\begin{align*} y &= 10(7-3) \\ &= 10(4) \\ &= 40 \\ \therefore \frac{dy}{dx} &= 0 \end{align*}$$

$$\begin{align*} q'(x) &= \frac{d}{dx}[x^4] - \frac{d}{dx}[6x^2] - \frac{d}{dx}[1] \\ &= 4x^3 - 6(2x) - 0 \\ &= 4x^3 - 12x \end{align*}$$

$$\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}[x^2] + \frac{d}{dx}[x] + \frac{d}{dx}[4] \\ &= 2x + 1 + 0 \\ &= 2x + 1 \end{align*}$$

$$\begin{align*} f'(x) &= \frac{1}{3} \cdot 3x^2 - 2x + 0 \\ &= x^2 - 2x \end{align*}$$

$$\begin{align*} \frac{dy}{dx} &= 3 \cdot \frac{3}{2}x^{\frac{3}{2}-1} - 4(1) + 0 \\ &= \frac{9}{2}x^{\frac{1}{2}} - 4 \\ &= \frac{9}{2}\sqrt{x} - 4 \end{align*}$$

First, expand $g(x)$:

$$\begin{align*} g(x) &= x(x+2) + 5x \\ &= x^2 + 2x + 5x \\ &= x^2 + 7x \end{align*}$$

Now differentiate:

$$\begin{align*} g'(x) &= 2x + 7 \end{align*}$$

$$\begin{align*} p(x) &= 200x^3 - 100x^2 + 40x - 8000 \\ p'(x) &= 200(3x^2) - 100(2x) + 40(1) - 0 \\ &= 600x^2 - 200x + 40 \end{align*}$$

First, expand the expression:

$$\begin{align*} y &= 14(x-1)\left(\frac{1}{2} + x^2\right) \\ &= 14\left(\frac{1}{2}x + x^3 - \frac{1}{2} - x^2\right) \\ &= 14x^3 - 14x^2 + 7x - 7 \end{align*}$$

Now differentiate:

$$\begin{align*} \frac{dy}{dx} &= 14(3x^2) - 14(2x) + 7(1) - 0 \\ &= 42x^2 - 28x + 7 \end{align*}$$

First, factor the numerator and simplify:

$$\begin{align*} f(x) &= \frac{x^2 - 5x + 6}{x - 2} \\ &= \frac{(x-2)(x-3)}{x-2} \\ &= x - 3 \quad \text{(for } x \neq 2 \text{)} \end{align*}$$

Now differentiate:

$$\begin{align*} f'(x) &= 1 \end{align*}$$

First, rewrite in exponential form:

$$\begin{align*} f(y) &= \sqrt{y} \\ &= y^{\frac{1}{2}} \end{align*}$$

Now differentiate:

$$\begin{align*} f'(y) &= \frac{1}{2}y^{\frac{1}{2}-1} \\ &= \frac{1}{2}y^{-\frac{1}{2}} \\ &= \frac{1}{2\sqrt{y}} \end{align*}$$

First, expand using difference of squares:

$$\begin{align*} f(z) &= (z-1)(z+1) \\ &= z^2 - 1 \end{align*}$$

Now differentiate:

$$\begin{align*} f'(z) &= 2z - 0 \\ &= 2z \end{align*}$$

First, divide each term by $x$:

$$\begin{align*} y &= \frac{x^3 + 2\sqrt{x} - 3}{x} \\ &= \frac{x^3}{x} + \frac{2\sqrt{x}}{x} - \frac{3}{x} \\ &= x^2 + 2x^{-\frac{1}{2}} - 3x^{-1} \end{align*}$$

Now differentiate:

$$\begin{align*} \frac{dy}{dx} &= 2x + 2\left(-\frac{1}{2}\right)x^{-\frac{3}{2}} - 3(-1)x^{-2} \\ &= 2x - x^{-\frac{3}{2}} + 3x^{-2} \\ &= 2x - \frac{1}{\sqrt{x^3}} + \frac{3}{x^2} \end{align*}$$

First, rewrite in exponential form:

$$\begin{align*} y &= \sqrt{x^3} + \frac{1}{3x^3} \\ &= x^{\frac{3}{2}} + \frac{1}{3}x^{-3} \end{align*}$$

Now differentiate:

$$\begin{align*} \frac{dy}{dx} &= \frac{3}{2}x^{\frac{3}{2}-1} + \frac{1}{3}(-3)x^{-3-1} \\ &= \frac{3}{2}x^{\frac{1}{2}} - x^{-4} \\ &= \frac{3}{2}\sqrt{x} - \frac{1}{x^4} \end{align*}$$

First, expand the square:

$$\begin{align*} \left[x^{\frac{3}{2}} - \frac{3}{x^{\frac{1}{2}}}\right]^2 &= \left[x^{\frac{3}{2}} - 3x^{-\frac{1}{2}}\right]^2 \\ &= \left(x^{\frac{3}{2}}\right)^2 - 2(x^{\frac{3}{2}})(3x^{-\frac{1}{2}}) + \left(3x^{-\frac{1}{2}}\right)^2 \\ &= x^3 - 6x + 9x^{-1} \end{align*}$$

Now differentiate:

$$\begin{align*} D_x\left[x^3 - 6x + 9x^{-1}\right] &= 3x^2 - 6 - 9x^{-2} \\ &= 3x^2 - 6 - \frac{9}{x^2} \end{align*}$$

First, make $y$ the subject of the formula:

$$\begin{align*} x &= 2y + 3 \\ x - 3 &= 2y \\ y &= \frac{1}{2}x - \frac{3}{2} \end{align*}$$

Now differentiate:

$$\begin{align*} \frac{dy}{dx} &= \frac{1}{2} \end{align*}$$

First, expand the square:

$$\begin{align*} f(\theta) &= 2(\theta^{\frac{3}{2}} - 3\theta^{-\frac{1}{2}})^2 \\ &= 2\left[(\theta^{\frac{3}{2}})^2 - 2(\theta^{\frac{3}{2}})(3\theta^{-\frac{1}{2}}) + (3\theta^{-\frac{1}{2}})^2\right] \\ &= 2(\theta^3 - 6\theta + 9\theta^{-1}) \\ &= 2\theta^3 - 12\theta + 18\theta^{-1} \end{align*}$$

Now differentiate:

$$\begin{align*} f'(\theta) &= 2(3\theta^2) - 12(1) + 18(-1)\theta^{-2} \\ &= 6\theta^2 - 12 - 18\theta^{-2} \\ &= 6\theta^2 - 12 - \frac{18}{\theta^2} \end{align*}$$

First, expand $(t+1)^3$:

$$\begin{align*} (t+1)^3 &= t^3 + 3t^2 + 3t + 1 \end{align*}$$

Now divide each term by $\sqrt{t} = t^{\frac{1}{2}}$:

$$\begin{align*} p(t) &= \frac{t^3 + 3t^2 + 3t + 1}{t^{\frac{1}{2}}} \\ &= t^{3-\frac{1}{2}} + 3t^{2-\frac{1}{2}} + 3t^{1-\frac{1}{2}} + t^{-\frac{1}{2}} \\ &= t^{\frac{5}{2}} + 3t^{\frac{3}{2}} + 3t^{\frac{1}{2}} + t^{-\frac{1}{2}} \end{align*}$$

Now differentiate:

$$\begin{align*} \frac{dp}{dt} &= \frac{5}{2}t^{\frac{3}{2}} + 3 \cdot \frac{3}{2}t^{\frac{1}{2}} + 3 \cdot \frac{1}{2}t^{-\frac{1}{2}} + (-\frac{1}{2})t^{-\frac{3}{2}} \\ &= \frac{5}{2}t^{\frac{3}{2}} + \frac{9}{2}t^{\frac{1}{2}} + \frac{3}{2}t^{-\frac{1}{2}} - \frac{1}{2}t^{-\frac{3}{2}} \\ &= \frac{5}{2}t\sqrt{t} + \frac{9}{2}\sqrt{t} + \frac{3}{2\sqrt{t}} - \frac{1}{2t\sqrt{t}} \end{align*}$$