Answer to Question #160451 in Calculus for effa

Question

find $y'$ for $y= x³ (sqrt(x² + 1))$

Answer 1

$y=x^3\cdot\sqrt{x²+1}$

$y'=(x^3)'\cdot\sqrt{x^2+1}+x^3\cdot(\sqrt{x^2+1})'=$

$\displaystyle 3x^2\cdot\sqrt{x^2+1}+x^3\cdot\frac{1}{2\sqrt{x^2+1}}\cdot(x^2+1)'=$

$\displaystyle 3x^2\sqrt{x^2+1}+x^3\cdot\frac{1}{2\sqrt{x^2+1}}\cdot2x=$

$\displaystyle 3x^2\sqrt{x^2+1}+\frac{x^4}{\sqrt{x^2+1}}=$

$\displaystyle 3x^2\sqrt{x^2+1}\cdot\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}+\frac{x^4}{\sqrt{x^2+1}}=$

$\displaystyle \frac{3x^2(x^2+1)}{\sqrt{x^2+1}}+\frac{x^4}{\sqrt{x^2+1}}=$

$\displaystyle \frac{3x^4+3x^2+x^4}{\sqrt{x^2+1}}=$

$\displaystyle \frac{4x^4+3x^2}{\sqrt{x^2+1}}$