Derivatives – Extra Practice

$$f'(x) = 7x^6$$

$$f'(x) = 20x^3 – 6x$$

Rewrite as \(x^{-3}\), then apply Power Rule:

$$f'(x) = -3x^{-4} = \frac{-3}{x^4}$$

$$f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$$

$$f'(x) = 4x^{-1/2} + 4x^{-2} = \frac{4}{\sqrt{x}} + \frac{4}{x^2}$$

$$f'(x) = 15x^4 – 21x^2 + 4x – 1$$

Let \(f = x^3\), \(g = \cos(x)\), so \(f’ = 3x^2\), \(g’ = -\sin(x)\)

$$f'(x) = 3x^2\cos(x) – x^3\sin(x)$$

Let \(f = 2x+5\), \(g = x^2-3\), so \(f’ = 2\), \(g’ = 2x\)

$$f'(x) = 2(x^2-3) + (2x+5)(2x) = 2x^2 – 6 + 4x^2 + 10x = 6x^2 + 10x – 6$$

Let \(f = x^4\), \(g = \ln(x)\), so \(f’ = 4x^3\), \(g’ = \frac{1}{x}\)

$$f'(x) = 4x^3\ln(x) + x^4 \cdot \frac{1}{x} = 4x^3\ln(x) + x^3 = x^3(4\ln(x) + 1)$$

Let \(f = e^x\), \(g = \sin(x)\), so \(f’ = e^x\), \(g’ = \cos(x)\)

$$f'(x) = e^x\sin(x) + e^x\cos(x) = e^x(\sin(x) + \cos(x))$$

Let \(f = x^2+1\), \(g = x^3-2x\), so \(f’ = 2x\), \(g’ = 3x^2-2\)

$$f'(x) = 2x(x^3-2x) + (x^2+1)(3x^2-2)$$

$$= 2x^4 – 4x^2 + 3x^4 – 2x^2 + 3x^2 – 2$$

$$= 5x^4 – 3x^2 – 2$$

Let \(f = x^3\), \(g = x+2\), so \(f’ = 3x^2\), \(g’ = 1\)

$$f'(x) = \frac{3x^2(x+2) – x^3(1)}{(x+2)^2} = \frac{3x^3 + 6x^2 – x^3}{(x+2)^2} = \frac{2x^3 + 6x^2}{(x+2)^2}$$

Let \(f = \cos(x)\), \(g = x^2\), so \(f’ = -\sin(x)\), \(g’ = 2x\)

$$f'(x) = \frac{-\sin(x) \cdot x^2 – \cos(x) \cdot 2x}{x^4}$$

$$= \frac{-x^2\sin(x) – 2x\cos(x)}{x^4} = \frac{-x\sin(x) – 2\cos(x)}{x^3}$$

Let \(f = e^x\), \(g = x^2+1\), so \(f’ = e^x\), \(g’ = 2x\)

$$f'(x) = \frac{e^x(x^2+1) – e^x(2x)}{(x^2+1)^2} = \frac{e^x(x^2 – 2x + 1)}{(x^2+1)^2} = \frac{e^x(x-1)^2}{(x^2+1)^2}$$

Let \(f = x^2-4\), \(g = x^2+4\), so \(f’ = 2x\), \(g’ = 2x\)

$$f'(x) = \frac{2x(x^2+4) – (x^2-4)(2x)}{(x^2+4)^2}$$

$$= \frac{2x^3 + 8x – 2x^3 + 8x}{(x^2+4)^2} = \frac{16x}{(x^2+4)^2}$$

$$f'(x) = 6(4x-1)^5 \cdot 4 = 24(4x-1)^5$$

Rewrite as \((3x^2+5)^{1/2}\):

$$f'(x) = \frac{1}{2}(3x^2+5)^{-1/2} \cdot 6x = \frac{6x}{2\sqrt{3x^2+5}} = \frac{3x}{\sqrt{3x^2+5}}$$

$$f'(x) = -\sin(4x^3) \cdot 12x^2 = -12x^2\sin(4x^3)$$

$$f'(x) = e^{x^2-3x} \cdot (2x – 3) = (2x-3)e^{x^2-3x}$$

$$f'(x) = \frac{1}{5x^3-2} \cdot 15x^2 = \frac{15x^2}{5x^3-2}$$

Outside: \((\cdot)^4\) → Inside: \(\frac{x+1}{x-1}\)

First find the derivative of the inside using Quotient Rule:

$$\frac{d}{dx}\left[\frac{x+1}{x-1}\right] = \frac{(x-1) – (x+1)}{(x-1)^2} = \frac{-2}{(x-1)^2}$$

Now apply Chain Rule:

$$f'(x) = 4\left(\frac{x+1}{x-1}\right)^3 \cdot \frac{-2}{(x-1)^2} = \frac{-8(x+1)^3}{(x-1)^5}$$

$$f'(x) = -5\sin(x) – 3\sec^2(x)$$

Rewrite as \([\sin(x)]^4\), then apply Chain Rule:

$$f'(x) = 4\sin^3(x) \cdot \cos(x)$$

$$f'(x) = \sec^2(3x^2+1) \cdot 6x = 6x\sec^2(3x^2+1)$$

Product Rule: \(f = x^2\), \(g = \sin(x)\)

$$f'(x) = 2x\sin(x) + x^2\cos(x)$$

Quotient Rule: \(f = \sin(x)\), \(g = 1+\cos(x)\)

$$f'(x) = \frac{\cos(x)(1+\cos(x)) – \sin(x)(-\sin(x))}{(1+\cos(x))^2}$$

$$= \frac{\cos(x) + \cos^2(x) + \sin^2(x)}{(1+\cos(x))^2}$$

$$= \frac{\cos(x) + 1}{(1+\cos(x))^2} = \frac{1}{1+\cos(x)}$$

$$f'(x) = 7e^{7x}$$

$$f'(x) = \frac{3x^2 + 4}{x^3 + 4x}$$

$$f'(x) = 5^x \ln(5)$$

Product Rule: \(f = x^3\), \(g = e^{2x}\)

$$f'(x) = 3x^2 e^{2x} + x^3 \cdot 2e^{2x} = e^{2x}(3x^2 + 2x^3) = x^2 e^{2x}(3 + 2x)$$

$$f'(x) = \frac{1}{\sin(x)} \cdot \cos(x) = \frac{\cos(x)}{\sin(x)} = \cot(x)$$

$$f'(x) = e^{\sin(x)} \cdot \cos(x) = \cos(x)e^{\sin(x)}$$

$$2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-x}{y}$$

$$3x^2 + 3y^2\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-x^2}{y^2}$$

Use Product Rule on the left side:

$$y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-y}{x}$$

Differentiate both sides:

$$2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$$

Collect \(\frac{dy}{dx}\) terms:

$$\frac{dy}{dx}(x^2 + 3y^2) = -2xy \Rightarrow \frac{dy}{dx} = \frac{-2xy}{x^2 + 3y^2}$$

$$\cos(y)\frac{dy}{dx} = 2x + \frac{dy}{dx}$$

$$\frac{dy}{dx}(\cos(y) – 1) = 2x \Rightarrow \frac{dy}{dx} = \frac{2x}{\cos(y) – 1}$$

Quotient Rule + Chain Rule:

$$f'(x) = \frac{3(x^2+1)^2 \cdot 2x \cdot e^x – (x^2+1)^3 \cdot e^x}{e^{2x}}$$

$$= \frac{e^x(x^2+1)^2\left[6x – (x^2+1)\right]}{e^{2x}}$$

$$= \frac{(x^2+1)^2(-x^2 + 6x – 1)}{e^x}$$

Recognize this is the Pythagorean identity: \(\sin^2(\theta) + \cos^2(\theta) = 1\)

So \(f(x) = 1\) for all x, therefore:

$$f'(x) = 0$$

Use log properties first: \(f(x) = \ln(x^2+1) – \ln(x-3)\)

$$f'(x) = \frac{2x}{x^2+1} – \frac{1}{x-3}$$

Product Rule + Chain Rule on both factors:

$$f'(x) = 4(x^2+3x)^3(2x+3) \cdot e^{2x} + (x^2+3x)^4 \cdot 2e^{2x}$$

$$= e^{2x}(x^2+3x)^3\left[4(2x+3) + 2(x^2+3x)\right]$$

$$= e^{2x}(x^2+3x)^3(2x^2 + 6x + 8x + 12)$$

$$= e^{2x}(x^2+3x)^3(2x^2 + 14x + 12)$$

$$= 2e^{2x}(x^2+3x)^3(x^2 + 7x + 6)$$

Differentiate both sides. Left side needs Chain Rule + Product Rule:

$$e^{xy}\!\left(y + x\frac{dy}{dx}\right) = 1 + \frac{dy}{dx}$$

$$ye^{xy} + xe^{xy}\frac{dy}{dx} = 1 + \frac{dy}{dx}$$

$$\frac{dy}{dx}(xe^{xy} – 1) = 1 – ye^{xy}$$

$$\frac{dy}{dx} = \frac{1 – ye^{xy}}{xe^{xy} – 1}$$