# How do you find the maclaurin series expansion of #e^(7x)ln((1-x)/3)#?

Note I've left the logs as

Taylor Expansion:

So Maclaurin series will be:

I'm only going to compute up to the second derivative, ie quadratic terms of x because this function will just get so messy it'll be easy to make a simple mistake. If you need more either work them out or use one of the various online tools.

Using product rule and chain rule:

Hence:

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To find the Maclaurin series expansion of ( e^{7x} \ln\left(\frac{1-x}{3}\right) ), we use the formula for the Maclaurin series expansion of a composite function. The formula states that if ( f(x) ) and ( g(x) ) have Maclaurin series expansions, then the Maclaurin series expansion of ( f(g(x)) ) is obtained by substituting the series expansion of ( g(x) ) into ( f(x) ), term by term.

Here, ( f(x) = e^x ) and ( g(x) = 7x \ln\left(\frac{1-x}{3}\right) ).

The Maclaurin series expansion of ( e^x ) is ( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} ).

For ( g(x) = 7x \ln\left(\frac{1-x}{3}\right) ), we need to find its Maclaurin series expansion. We'll use the properties of logarithms and the geometric series to expand ( \ln\left(\frac{1-x}{3}\right) ).

( \ln\left(\frac{1-x}{3}\right) = \ln(1-x) - \ln(3) )

( = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots\right) - \ln(3) )

Now, multiply by ( 7x ):

( 7x \ln\left(\frac{1-x}{3}\right) = -7x^2 - \frac{7x^3}{2} - \frac{7x^4}{3} - \ldots - 7x\ln(3) )

Now, substitute this series expansion into the Maclaurin series expansion of ( e^{7x} ):

( e^{7x} \ln\left(\frac{1-x}{3}\right) = e^{7x}(-7x^2 - \frac{7x^3}{2} - \frac{7x^4}{3} - \ldots - 7x\ln(3)) )

This gives us the Maclaurin series expansion of ( e^{7x} \ln\left(\frac{1-x}{3}\right) ).

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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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