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Question: differentiate: x^5(3-6x^-9) by first principle...

differentiate: x^5(3-6x^-9) by first principle

Answer

15x^4 + 24x^{-5}

Explanation

Solution

To differentiate the given function f(x)=x5(3−6x−9)f(x) = x^5(3-6x^{-9}) by the first principle, we follow these steps:

1. Simplify the function: First, expand and simplify the given function: f(x)=x5(3−6x−9)f(x) = x^5(3-6x^{-9}) f(x)=3x5−6x5⋅x−9f(x) = 3x^5 - 6x^5 \cdot x^{-9} Using the rule am⋅an=am+na^m \cdot a^n = a^{m+n}: f(x)=3x5−6x5−9f(x) = 3x^5 - 6x^{5-9} f(x)=3x5−6x−4f(x) = 3x^5 - 6x^{-4}

2. Apply the definition of differentiation by first principle: The first principle states that the derivative of a function f(x)f(x) is given by: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

3. Find f(x+h)f(x+h): Substitute (x+h)(x+h) into the simplified function f(x)=3x5−6x−4f(x) = 3x^5 - 6x^{-4}: f(x+h)=3(x+h)5−6(x+h)−4f(x+h) = 3(x+h)^5 - 6(x+h)^{-4}

4. Substitute f(x+h)f(x+h) and f(x)f(x) into the first principle formula: f′(x)=lim⁡h→0[3(x+h)5−6(x+h)−4]−[3x5−6x−4]hf'(x) = \lim_{h \to 0} \frac{[3(x+h)^5 - 6(x+h)^{-4}] - [3x^5 - 6x^{-4}]}{h} f′(x)=lim⁡h→03(x+h)5−6(x+h)−4−3x5+6x−4hf'(x) = \lim_{h \to 0} \frac{3(x+h)^5 - 6(x+h)^{-4} - 3x^5 + 6x^{-4}}{h} Rearrange the terms: f′(x)=lim⁡h→03[(x+h)5−x5]−6[(x+h)−4−x−4]hf'(x) = \lim_{h \to 0} \frac{3[(x+h)^5 - x^5] - 6[(x+h)^{-4} - x^{-4}]}{h}

5. Split the limit and apply the standard limit formula: We can split the limit into two parts: f′(x)=lim⁡h→0(3(x+h)5−x5h−6(x+h)−4−x−4h)f'(x) = \lim_{h \to 0} \left( 3 \frac{(x+h)^5 - x^5}{h} - 6 \frac{(x+h)^{-4} - x^{-4}}{h} \right) We use the standard limit formula: lim⁡h→0(x+h)n−xnh=nxn−1\lim_{h \to 0} \frac{(x+h)^n - x^n}{h} = nx^{n-1}.

For the first term (n=5n=5): lim⁡h→03(x+h)5−x5h=3⋅(5x5−1)=3⋅5x4=15x4\lim_{h \to 0} 3 \frac{(x+h)^5 - x^5}{h} = 3 \cdot (5x^{5-1}) = 3 \cdot 5x^4 = 15x^4

For the second term (n=−4n=-4): lim⁡h→0−6(x+h)−4−x−4h=−6⋅(−4x−4−1)=−6⋅(−4x−5)=24x−5\lim_{h \to 0} -6 \frac{(x+h)^{-4} - x^{-4}}{h} = -6 \cdot (-4x^{-4-1}) = -6 \cdot (-4x^{-5}) = 24x^{-5}

6. Combine the results: Add the results from both terms to get the final derivative: f′(x)=15x4+24x−5f'(x) = 15x^4 + 24x^{-5}

The derivative can also be written as 15x4+24x515x^4 + \frac{24}{x^5}.

The final answer is 15x4+24x−515x^4 + 24x^{-5}.

Explanation of the solution:

  1. Simplify the given function f(x)=x5(3−6x−9)f(x) = x^5(3-6x^{-9}) to f(x)=3x5−6x−4f(x) = 3x^5 - 6x^{-4}.
  2. Use the first principle definition: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
  3. Substitute f(x+h)=3(x+h)5−6(x+h)−4f(x+h) = 3(x+h)^5 - 6(x+h)^{-4} and f(x)f(x) into the formula.
  4. Separate the terms: 3lim⁡h→0(x+h)5−x5h−6lim⁡h→0(x+h)−4−x−4h3 \lim_{h \to 0} \frac{(x+h)^5 - x^5}{h} - 6 \lim_{h \to 0} \frac{(x+h)^{-4} - x^{-4}}{h}.
  5. Apply the standard limit formula lim⁡h→0(x+h)n−xnh=nxn−1\lim_{h \to 0} \frac{(x+h)^n - x^n}{h} = nx^{n-1} for each term.
  6. For the first term, n=5n=5, resulting in 3⋅5x5−1=15x43 \cdot 5x^{5-1} = 15x^4.
  7. For the second term, n=−4n=-4, resulting in −6⋅(−4)x−4−1=24x−5-6 \cdot (-4)x^{-4-1} = 24x^{-5}.
  8. Combine the results: 15x4+24x−515x^4 + 24x^{-5}.