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)+f'(0)x+1/2 f''(0)x^2;
Which here gives f(x)=4^(-2) - 6*(4)^(-3) x + 27*(4)^(-4) x^2.