Chapter 2: Q. B (page 238)

Use the definition of the derivative and factoring formulas to prove that for any positive integer k, the derivative of $x^{k}$ is $k x^{k - 1}$

Short Answer

Expert verified

$\frac{d}{d x} (x^{k}) = k x^{k - 1}$

Step by step solution

Step 1:Given information

The given expresiion is $f (x) = x^{k}$

Step 2:Explanation

We’ll need to restrict $k$ to be a positive integer. In this case if we define $f (x) = x^{k}$ ,

we know from the alternate limit form of the definition of the derivative that the derivative $f' (a)$ is given by

$f' (a) = \lim_{x \to a} \frac{f (x) - f (a)}{x - a} = \frac{x^{k} - x^{a}}{x - a}$

Now,

$x^{k} - x^{a} = (x - a) (x^{k - 1} + a x^{k - 2} + . . . a^{k - 2} x + a^{k - 1})$

If we put this into the formula for the derivative we see that we can cancel the $(x - a)$ and then compute the limit.

$f' (a) = \lim_{x \to a} \frac{(x - a) (x^{k - 1} + a x^{k - 2} + . . . a^{k - 2} x + a^{k - 1})}{x - a}$

$\Rightarrow f' (a) = \lim_{x \to a} (x^{k - 1} + a x^{k - 2} + . . . a^{k - 2} x + a^{k - 1})$

$\Rightarrow f' (a) = k a^{k - 1}$

To completely finish this off we simply replace the $a$ with an $x$ to get,

$\Rightarrow f' (x) = k x^{k - 1}$

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Use the definition of the derivative and factoring formulas to prove that for any positive integer k, the derivative of $x^{k}$ is $k x^{k - 1}$

Short Answer

Step by step solution

Step 1:Given information

Step 2:Explanation

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Use the definition of the derivative and factoring formulas to prove that for any positive integer k, the derivative ofxkis kxk-1

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Use the definition of the derivative and factoring formulas to prove that for any positive integer k, the derivative of $x^{k}$ is $k x^{k - 1}$