Question

derivada y = log 10 (3x²+2)^5 (log na base 10)

Answer 1

Explicação passo-a-passo:

Temos que:

$\sf y=log_{a}~u~\Rightarrow~y'=\dfrac{u'}{u\cdot ln~a}$

Assim:

$\sf y=log_{10}~(3x^2+2)^5$

$\sf y'=\dfrac{5\cdot(3x^2+2)^4\cdot(3x^2+2)'}{(3x^2+2)^5\cdot ln~10}$

$\sf y'=\dfrac{5\cdot2\cdot3x}{(3x^2+2)\cdot ln~10}$

$\sf y'=\dfrac{30x}{(3x^2+2)\cdot ln~10}$