Question

Quanto que é a derivada da função ?
$y= ( x^{2} +1) ^{3} .ln 3x$

Answer 1

$y(x)=\left(x^2+1\right)^{3}\cdot\ln(3x)\implies \\u(v)=(v)^3\\v(x)=x^2+1\\w(t)=\ln t\\t(x)=3x\\\\\displaystyle \frac{dy}{dx}=\frac{du}{dx}\cdot w+\frac{dw}{dx}\cdot u\\\\\frac{du}{dx}=\frac{d}{dv}v^3\cdot\frac{d}{dx}x^2+1=3v^2\cdot2x=\underline{6x(x^2+1)^2}\\\\ \frac{dw}{dx}=\frac{d}{dt}\ln t\cdot\frac{d}{dx}3x=\frac{1}{t}\cdot 3=\frac{3}{3x}=\underline{\frac{1}{x}}$
$\displaystyle \\\\\frac{dy}{dx}=\left(6x(x^2+1)^2\right)\cdot\ln(3x)+\frac{1}{x}\cdot(x^2+1)^3=\\\\\frac{x\left(6x(x^2+1)^2\ln(3x)\right)}{x}+\frac{(x^2+1)^2}{x}=\\\\\\\boxed{\frac{\left(6x^2(x^2+1)^2\ln(3x)\right)+(x^2+1)^2}{x}}$