Question

find the value of dy/dx if y = log (
$3x ^{2} + 2x + 1$
​

Answer 1

Answer:

it's very easy and I hope it helps u

Answer 2

$\boxed{\huge\red{ \boxed{\mathfrak{solution}}}}$

To find dy/dx of :

$y = log(3 {x}^{2} + 2x + 1) \\ \\ \rightarrow \: \frac{dy}{dx} = \frac{d}{dx} ( log(3 {x}^{2} + 2x + 1 )) \times \frac{d}{dx} (3 {x}^{2} + 2x + 1) \\ \\ \rightarrow \: \frac{dy}{dx} = \frac{1}{3 {x}^{2} + 2x + 1 } \times (6x + 2) \\ \\ \rightarrow \: \frac{dy}{dx} = \frac{6x + 2}{3 {x}^{2} + 2x + 1}$

Formula used :

•

$\boxed{\bold{ \purple{ \frac{d}{dx} log(x) = \frac{1}{x}} }}$

•

$\boxed{ \bold{ \blue{\frac{d}{dx} {x}^{y} = y {x}^{y - 1} }}}$