How do you use differentiation to find a power series representation for #1/(6+x)^2#?
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To find the power series representation for ( \frac{1}{{(6+x)}^2} ), we can use differentiation. Here's the stepbystep process:

Start with a known power series representation: [ \frac{1}{{1x}} = \sum_{n=0}^{\infty} x^n ]

Differentiate both sides with respect to ( x ): [ \frac{d}{{dx}} \left( \frac{1}{{1x}} \right) = \frac{d}{{dx}} \left( \sum_{n=0}^{\infty} x^n \right) ]

Use the power rule for differentiation and recognize that the derivative of a power series is just another power series: [ \frac{1}{{(1x)^2}} = \sum_{n=1}^{\infty} nx^{n1} ]

Now, to find the power series representation for ( \frac{1}{{(6+x)}^2} ), substitute ( x = \frac{x}{6} ) into the series above: [ \frac{1}{{(6+x)}^2} = \sum_{n=1}^{\infty} n \left( \frac{x}{6} \right)^{n1} ]

To adjust the index of summation, let ( m = n1 ), then rewrite the series: [ \frac{1}{{(6+x)}^2} = \sum_{m=0}^{\infty} (m+1) \left( \frac{x}{6} \right)^m ]

Simplify the series: [ \frac{1}{{(6+x)}^2} = \sum_{m=0}^{\infty} (1)^m \frac{(m+1)x^m}{6^m} ]
Therefore, the power series representation for ( \frac{1}{{(6+x)}^2} ) is: [ \sum_{m=0}^{\infty} (1)^m \frac{(m+1)x^m}{6^m} ]
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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