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For f(x) = (3x+4)^(-2), find f'(x) and f''(x) and hence write down the Maclaurin series up to and including the term in x^2.

f'(x)=-2(3x+4)^(-3) * 3 = -6(3x+4)^(-3);
f''(x)= 18(3x+4)^(-4) * 3 = 54(3x+4)^(-4);
both found by using the chain rule for differentiation.

Then Maclaurin series up to x^2 is: f(x)=f(0)...

JM
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