Specialization Math Example 4

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Example 4

medium
The general derivative rule is (xn)′=nxn−1(x^n)' = nx^{n-1}. Specialise to find the derivatives of x3x^3, x1/2x^{1/2}, and x−1x^{-1}.

Solution

  1. 1
    n=3n=3: (x3)′=3x2(x^3)' = 3x^2.
  2. 2
    n=12n=\frac{1}{2}: (x1/2)′=12x−1/2=12x(x^{1/2})' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.
  3. 3
    n=−1n=-1: (x−1)′=−1⋅x−2=−1x2(x^{-1})' = -1 \cdot x^{-2} = -\frac{1}{x^2}.

Answer

(x3)′=3x2,(x)′=12x,(x−1)′=−1x2(x^3)'=3x^2,\quad (\sqrt{x})'=\frac{1}{2\sqrt{x}},\quad (x^{-1})'=-\frac{1}{x^2}
The power rule is a general formula. Specialising to particular values of nn (integer, fraction, negative) produces standard derivative results used throughout calculus.

About Specialization

Applying a general theorem or formula to a specific case by substituting particular values for the variables or parameters.

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Example 1 easy

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Example 2 medium

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Example 3 easy

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