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The following results by product rule and quotient rule are listed below:

  1. [tex]f'(x) = 3\cdot x^{2}-15\cdot x - 9[/tex]
  2. [tex]f'(x) = \frac{15}{2} \cdot x^{4}-6\cdot x^{3}+\frac{45}{2}\cdot x^{2}-\frac{81}{2} \cdot x +13[/tex]
  3. [tex]f'(x) = \frac{90\cdot x^{2}-45\cdot x^{2}-40\cdot x +72}{25\cdot x^{4}-80\cdot x^{2}+64}[/tex]
  4. [tex]f'(x) = \frac{\left(9\cdot x^{2}-18\cdot x-6\right)\cdot \left(\frac{x^{2}}{2}+7\cdot x-8 \right)-\left(3\cdot x^{3}-9\cdot x^{2}-6\cdot x +1 \right)\cdot \left(\frac{2}{3}\cdot x+7 \right)}{\left(\frac{x^{2}}{2}+7\cdot x-8 \right)^{2}}[/tex]

The following results by chain rule are listed below:

  1. [tex]f'(x) = 3\cdot (2\cdot x^{2}-2\cdot x +1)^{2}\cdot (4\cdot x-2)[/tex]
  2. [tex]f'(x) = \frac{15\cdot x^{2}-16\cdot x+9}{2\cdot \sqrt{5\cdot x^{3}-8\cdot x^{2}+9\cdot x-6}}[/tex]
  3. [tex]f'(x) = \frac{2\cdot \left(6\cdot x-\frac{2}{3}-4\cdot x^{-5} \right)}{3\cdot \sqrt[3]{3\cdot x^{2}-\frac{2}{3}\cdot x +x^{-4} } }[/tex]

How to apply product rule and quotient rule to obtain the derivative of function

In this part we must apply the following rules to calculate the derivatives:

Product rule

[tex]\frac{d}{dx}[f(x)\cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)[/tex]   (1)

Quotient rule

[tex]\frac{d}{dx}\left[\frac{f(x)}{g(x)} \right] = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^{2}}[/tex]   (2)

Now we proceed to obtain each expression:

1) [tex]f'(x) = 3\cdot x\cdot (x-6)+3\cdot x-9[/tex]

[tex]f'(x) = 3\cdot x^{2}-15\cdot x - 9[/tex] [tex]\blacksquare[/tex]

2) [tex]f'(x) = \left(\frac{9}{4}\cdot x^{2}-4\cdot x+\frac{1}{4}\right)\cdot (2\cdot x^{2}-2\cdot x + 4) + \left(\frac{3}{4}\cdot x^{3}-2\cdot x^{2}+\frac{x}{4}-6\right)\cdot (4\cdot x-2)[/tex]

[tex]f'(x) = \frac{9}{2}\cdot x^{4}-8\cdot x^{3}+\frac{1}{2}\cdot x^{2}-\frac{9}{2}\cdot x^{3}+8\cdot x^{2}-\frac{x}{2}+9\cdot x^{2}-16\cdot x+1 +3\cdot x^{4}-8\cdot x^{3}+x^{2}-24\cdot x-\frac{3}{2}\cdot x^{3}+4\cdot x^{2}-\frac{x}{2} +12[/tex]

[tex]f'(x) = \left(\frac{9}{2}+3 \right)\cdot x^{4}+\left(-8-\frac{9}{2}+8-\frac{3}{2} \right)\cdot x^{3} + \left(\frac{1}{2}+8+9+1+4 \right)\cdot x^{2} + \left(-16-24-\frac{1}{2} \right)\cdot x + \left(1+12\right)[/tex]

[tex]f'(x) = \frac{15}{2} \cdot x^{4}-6\cdot x^{3}+\frac{45}{2}\cdot x^{2}-\frac{81}{2} \cdot x +13[/tex]   [tex]\blacksquare[/tex]

3) [tex]f'(x) = \frac{-9\cdot (5\cdot x^{2}-8)-(-9\cdot x+4)\cdot (10\cdot x)}{(5\cdot x^{2}-8)^{2}}[/tex]

[tex]f'(x) = \frac{-45\cdot x^{2}+72+90\cdot x^{2}-40\cdot x}{25\cdot x^{4}-80\cdot x^{2}+64}[/tex]

[tex]f'(x) = \frac{90\cdot x^{2}-45\cdot x^{2}-40\cdot x +72}{25\cdot x^{4}-80\cdot x^{2}+64}[/tex]   [tex]\blacksquare[/tex]

4) [tex]f'(x) = \frac{\left(9\cdot x^{2}-18\cdot x-6\right)\cdot \left(\frac{x^{2}}{2}+7\cdot x-8 \right)-\left(3\cdot x^{3}-9\cdot x^{2}-6\cdot x +1 \right)\cdot \left(\frac{2}{3}\cdot x+7 \right)}{\left(\frac{x^{2}}{2}+7\cdot x-8 \right)^{2}}[/tex]   [tex]\blacksquare[/tex]

How to apply chain rule to find derivatives

Let be a function of the form [tex]f[u(x)][/tex], the derivative of this function is defined by the following rule:

[tex]\frac{d}{dx}[f[u(x)]] = \frac{df}{du}\cdot \frac{du}{dx}[/tex]

Now we proceed to present the following results:

1) [tex]f'(x) = 3\cdot (2\cdot x^{2}-2\cdot x +1)^{2}\cdot (4\cdot x-2)[/tex]   [tex]\blacksquare[/tex]

2) [tex]f'(x) = \frac{15\cdot x^{2}-16\cdot x+9}{2\cdot \sqrt{5\cdot x^{3}-8\cdot x^{2}+9\cdot x-6}}[/tex]   [tex]\blacksquare[/tex]

3) [tex]f'(x) = \frac{2}{3}\cdot \left(3\cdot x^{2}-\frac{2}{3}\cdot x+x^{-4} \right)^{-\frac{1}{3} } \cdot \left(6\cdot x-\frac{2}{3} -4\cdot x^{-5} \right)[/tex]   [tex]\blacksquare[/tex]

[tex]f'(x) = \frac{2\cdot \left(6\cdot x-\frac{2}{3}-4\cdot x^{-5} \right)}{3\cdot \sqrt[3]{3\cdot x^{2}-\frac{2}{3}\cdot x +x^{-4} } }[/tex]   [tex]\blacksquare[/tex]

To learn more on derivatives, we kindly invite to check this verified question: https://brainly.com/question/23847661

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