Calculus
f′(x)=2x(3+1x)2x−1f^{\prime}\left(x\right)=2x\left(3+\frac{1}{x}\right)^{2x-1}
y′=2(3+1x)2x−1((3x+1)ln(3+1x)−1)xy^{\prime}=\frac{2\left(3+\frac{1}{x}\right)^{^{2x-1}}\left(\left(3x+1\right)\ln\left(3+\frac{1}{x}\right)-1\right)}{x}
y′=2(3+1x)2x−1(ln(3+1x)−1)3x+1y^{\prime}=\frac{2\left(3+\frac{1}{x}\right)^{^{2x-1}}\left(\ln\left(3+\frac{1}{x}\right)-1\right)}{3x+1}
f′(x)=2(3+1x)2xln(3+1x)+2(3+1x)2x−2f^{\prime}\left(x\right)=2\left(3+\frac{1}{x}\right)^{2x}\ln\left(3+\frac{1}{x}\right)+2\left(3+\frac{1}{x}\right)^{2x-2}