Question

Evaluate the following derivatives. \(\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\left({x}+{1}\right)}^{{{2}{x}}}\right)}\)

Answer 1

Step 1
To Determine: Evaluate the following derivatives.
Given: we have a function \(\displaystyle{\left({x}+{1}\right)}^{{2}}{x}\)
Explanation, we have a function \(\displaystyle{y}={\left({x}+{1}\right)}^{{2}}{x}\) we have to find the derivative of y now we will applying exponent rule \(\displaystyle{a}^{{b}}={e}^{{{b}{\ln{{\left({a}\right)}}}}}{s}{o}{w}{e}{h}{a}{v}{e}{\left({x}+{1}\right)}^{{2}}{x}={e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}\) now we will applying the chain rule
\(\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}\right)}={e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({2}{x}{\ln{{\left({x}+{1}\right)}}}\right)}\)
Step 2
Now we will use product rule
\(\displaystyle{e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({2}{x}{\ln{{\left({x}+{1}\right)}}}\right)}={e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}\times{2}\times{\left[{x}\frac{{d}}{{{\left.{d}{x}\right.}}}{\ln{{\left({x}+{1}\right)}}}+{\ln{{\left({x}+{1}\right)}}}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{x}\right.}}}\right]}\)
\(\displaystyle={2}{e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}{\left[\frac{{x}}{{{x}+{1}}}+{\ln{{\left({x}+{1}\right)}}}\right]}\)
\(\displaystyle={2}{\left({x}+{1}\right)}^{{{2}{x}}}{\left[\frac{{x}}{{{x}+{1}}}+{\ln{{\left({x}+{1}\right)}}}\right]}\)