Question

# Find the derivatives of the functions. u(x)=\frac{3^{x^{2}}}{x^{2}+1}

Derivatives
Find the derivatives of the functions.
$$\displaystyle{u}{\left({x}\right)}={\frac{{{3}^{{{x}^{{{2}}}}}}}{{{x}^{{{2}}}+{1}}}}$$

2021-06-24

$$\displaystyle{u}{\left({x}\right)}={\frac{{{3}^{{{x}^{{{2}}}}}}}{{{x}^{{{2}}}+{1}}}}$$ Differentiate both sides with respect to x
$$\displaystyle{u}'{\left({x}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\frac{{{3}^{{{x}^{{{2}}}}}}}{{{x}^{{{2}}}+{1}}}}\right]}$$
Apply the Quotient Rule
$$\displaystyle{u}'{\left({x}\right)}={\left({x}^{{{2}}}+{1}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{3}^{{{x}^{{{2}}}}}\right]}-{3}{x}^{{{2}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{x}^{{{2}}}+{1}\right]}$$
Apply $$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{b}^{{{u}}}\right]}={b}^{{{u}}}{\ln{{b}}}{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}$$ and $$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{x}^{{{n}}}\right]}={n}{x}^{{{n}-{1}}}$$
$$u'(x)=\frac{(x^{2}+1)(3^{x^{2}})(\ln3)\frac{d}{dx}[x^{2}]-3^{x^{2}}(2x^{2-1})}{(x^{2}+1)^{2}}$$
Simplify
$$\displaystyle{u}'{\left({x}\right)}={\frac{{{\left({x}^{{{2}}}+{1}\right)}{\left({3}^{{{x}^{{{2}}}}}\right)}{\left({\ln{{3}}}\right)}{\left({2}{x}\right)}-{3}^{{{x}^{{2}}}}{\left({2}{x}\right)}}}{{{\left({x}^{{{2}}}+{1}\right)}^{{{2}}}}}}$$
Factor out $$\displaystyle{2}{x}\cdot{3}^{{{x}^{{{2}}}}}$$
$$\displaystyle{u}'{\left({x}\right)}={\frac{{{2}{x}\cdot{3}^{{{x}^{{{2}}}}}{\left[{\left({x}^{{{2}}}+{1}\right)}{\ln{{3}}}-{x}\right]}}}{{{\left({x}^{{{2}}}+{1}\right)}^{{{2}}}}}}$$