How do you take the derivative of d/dx [3x(x^2+1)^3] using the chain rule?
\[\huge \frac{d}{dx}[3x(x^2+1)^3] \] this problem?
yes
\[\huge \frac{d}{dx}[3x(x^2+1)^3] \] \[\huge =3\frac{d}{dx}[x(x^2+1)^3] \] \[\huge =3\cdot([x]'(x^2+1)^3+x[(x^2+1)^3]') \] \[\huge =3\cdot(1\cdot(x^2+1)^3+x\cdot 3(x^2+1)^2\cdot [x^2+1]') \] can you take it from here?
that last line should read: \[\large =3\cdot(1\cdot(x^2+1)^3+x\cdot 3(x^2+1)^2\cdot [x^2+1]') \]
okay thanks. let me see if I can work it out.
ok...:) tag me if you still need help...
okay i got as far as 3(x^2+1)^3 + 18x^2(x^2+1)^2 but my book says the answer is 3(x^2+1)^2(7x^2+1). Im confused as to how it simplifies to that final answer
\(\large =3\cdot(1\cdot(x^2+1)^3+x\cdot 3(x^2+1)^2\cdot [x^2+1]') \) \(\large =3\cdot(1\cdot(x^2+1)^3+x\cdot 3(x^2+1)^2\cdot (2x) \) \(\large =3\cdot((x^2+1)^3+6x^2(x^2+1)^2) \) \(\large =3(x^2+1)^2((x^2+1)+6x^2) \) \(\large =3(x^2+1)^2(7x^2+1) \)
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