# How do you find the power series for #f'(x)# and #int f(t)dt# from [0,x] given the function #f(x)=Sigma sqrtnsqrt(n+1)x^n# from #n=[1,oo)#?

for

using the ratio test we have:

so:

then:

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To find the power series for ( f'(x) ), differentiate each term of the given series for ( f(x) ) and then rewrite the series. To find ( \int f(t)dt ) from ( [0,x] ), integrate each term of the given series for ( f(x) ) with respect to ( t ) and then rewrite the series.

For ( f'(x) ), differentiate each term of the series for ( f(x) ) with respect to ( x ), term by term:

[ f'(x) = \sum_{n=1}^{\infty} \sqrt{n\sqrt{n+1}} \cdot n \cdot x^{n-1} ]

For ( \int f(t) dt ) from ( [0,x] ), integrate each term of the series for ( f(x) ) with respect to ( t ), term by term:

[ \int f(t) dt = \sum_{n=1}^{\infty} \frac{\sqrt{n\sqrt{n+1}}}{n+1} \cdot x^{n+1} + C ]

where ( C ) is the constant of integration.

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When evaluating a one-sided 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 one-sided 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 one-sided 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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