Hope Hancock

2022-10-03

Say$f(x)=\mathrm{ln}(1+2x+2{x}^{2})$ or $g(x)=\mathrm{tan}(2{x}^{4}-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of $\frac{1}{1+x}$ or $\frac{1}{1-x}$ and then manipulate it to get result

trutdelamodej0

Beginner2022-10-04Added 11 answers

keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,

$\frac{1}{1+t}=1-t+{t}^{2}-{t}^{3}+{t}^{4}-{t}^{5}\cdots $

$\mathrm{log}(1+t)=t-\frac{{t}^{2}}{2}+\frac{{t}^{3}}{3}-\frac{{t}^{4}}{4}\cdots $

Taking $t=2x+2{x}^{2}$ correctly gives the first few terms of $\mathrm{log}(1+2x+2{x}^{2}),$, up to ${x}^{4}$

$\mathrm{log}(1+2x+2{x}^{2})=2x-\frac{4{x}^{3}}{3}+2{x}^{4}\cdots $

$\frac{1}{1+t}=1-t+{t}^{2}-{t}^{3}+{t}^{4}-{t}^{5}\cdots $

$\mathrm{log}(1+t)=t-\frac{{t}^{2}}{2}+\frac{{t}^{3}}{3}-\frac{{t}^{4}}{4}\cdots $

Taking $t=2x+2{x}^{2}$ correctly gives the first few terms of $\mathrm{log}(1+2x+2{x}^{2}),$, up to ${x}^{4}$

$\mathrm{log}(1+2x+2{x}^{2})=2x-\frac{4{x}^{3}}{3}+2{x}^{4}\cdots $

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