# Evaluate the following derivatives. d/(dx)((x+1)^(2x))

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

2021-02-20
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
$$\displaystyle{e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({2}{x}{\ln{{\left({x}+{1}\right)}}}\right)}={e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}\times{2}\times{\left[{x}\frac{{d}}{{{\left.{d}{x}\right.}}}{\ln{{\left({x}+{1}\right)}}}+{\ln{{\left({x}+{1}\right)}}}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{x}\right.}}}\right]}$$
$$\displaystyle={2}{e}^{{{2}{x}{\ln{{\left({x}+{1}\right)}}}}}{\left[\frac{{x}}{{{x}+{1}}}+{\ln{{\left({x}+{1}\right)}}}\right]}$$
$$\displaystyle={2}{\left({x}+{1}\right)}^{{{2}{x}}}{\left[\frac{{x}}{{{x}+{1}}}+{\ln{{\left({x}+{1}\right)}}}\right]}$$

### Relevant Questions

Evaluate the following derivatives.
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{-{x}}}^{{{x}}}}{\frac{{{\left.{d}{t}\right.}}}{{{t}^{{{10}}}+{1}}}}$$
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Evaluate the following derivatives.
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$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{9}}}{{{x}^{{{5}}}}}}\right)}$$
Find the following higher-order derivatives.
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$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{{1}}}^{{{x}}}}{\ln{{t}}}{\left.{d}{t}\right.}$$