Math, asked by yajahiratorres200, 10 months ago

d/dx [log 2 (2x²+1)²] ayuda por favor

Answers

Answered by muthyalasravani1729

0

Answer:

8x/(2x^2+1)

Step-by-step explanation:

log[2(2x^2+1)^2]=log2+log[2x^2+1]^2

=log2+2log(2x^2+1)

differential with respect to x

0+2[(1/(2x^2+1))4x]

8x/(2x^2+1)

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