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}