Using the 1st principle of differentiation find the derivative of X²

Mathematics

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Answer

−2x3-\frac{2}{x^3}

Step-by-step solution

5 steps

To find the derivative of f(x)=1x2f(x) = \frac{1}{x^2} using the first principle, we use the definition:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  1. Identify f(x)f(x) and f(x+h)f(x+h).

    Given f(x)=1x2f(x) = \frac{1}{x^2}.
    Then f(x+h)=1(x+h)2f(x+h) = \frac{1}{(x+h)^2}.

  2. Substitute f(x)f(x) and f(x+h)f(x+h) into the first principle formula.

    f′(x)=lim⁡h→01(x+h)2−1x2hf'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}

  3. Simplify the numerator by finding a common denominator.

    f′(x)=lim⁡h→0x2−(x+h)2x2(x+h)2hf'(x) = \lim_{h \to 0} \frac{\frac{x^2 - (x+h)^2}{x^2(x+h)^2}}{h} f′(x)=lim⁡h→0x2−(x2+2xh+h2)hx2(x+h)2f'(x) = \lim_{h \to 0} \frac{x^2 - (x^2 + 2xh + h^2)}{h x^2(x+h)^2} f′(x)=lim⁡h→0x2−x2−2xh−h2hx2(x+h)2f'(x) = \lim_{h \to 0} \frac{x^2 - x^2 - 2xh - h^2}{h x^2(x+h)^2} f′(x)=lim⁡h→0−2xh−h2hx2(x+h)2f'(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h x^2(x+h)^2}

  4. Factor out hh from the numerator and cancel it with the hh in the denominator.

    f′(x)=lim⁡h→0h(−2x−h)hx2(x+h)2f'(x) = \lim_{h \to 0} \frac{h(-2x - h)}{h x^2(x+h)^2} f′(x)=lim⁡h→0−2x−hx2(x+h)2f'(x) = \lim_{h \to 0} \frac{-2x - h}{x^2(x+h)^2}

  5. Evaluate the limit by substituting h=0h=0.

    f′(x)=−2x−0x2(x+0)2f'(x) = \frac{-2x - 0}{x^2(x+0)^2} f′(x)=−2xx2(x2)f'(x) = \frac{-2x}{x^2(x^2)} f′(x)=−2xx4f'(x) = \frac{-2x}{x^4} f′(x)=−2x3f'(x) = -\frac{2}{x^3}

    The derivative of 1x2\frac{1}{x^2} using the first principle is:

    −2x3-\frac{2}{x^3}
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