Question

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

Derivatives
ANSWERED
asked 2021-06-23
Find the derivatives of the functions.
\(\displaystyle{u}{\left({x}\right)}={\frac{{{3}^{{{x}^{{{2}}}}}}}{{{x}^{{{2}}}+{1}}}}\)

Answers (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 \(\displaystyle{2}{x}\cdot{3}^{{{x}^{{{2}}}}}\)
\(\displaystyle{u}'{\left({x}\right)}={\frac{{{2}{x}\cdot{3}^{{{x}^{{{2}}}}}{\left[{\left({x}^{{{2}}}+{1}\right)}{\ln{{3}}}-{x}\right]}}}{{{\left({x}^{{{2}}}+{1}\right)}^{{{2}}}}}}\)

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