Differentiate the following function respect to x p=e^{x^{2}\sin x}

luipieduq3 2021-11-22 Answered
Differentiate the following function respect to x
\(\displaystyle{p}={e}^{{{x}^{{{2}}}{\sin{{x}}}}}\)

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Expert Answer

Sue Leahy
Answered 2021-11-23 Author has 28 answers
Step 1
Given that:
\(\displaystyle{p}={e}^{{{x}^{{{2}}}{\sin{{x}}}}}\)
Step 2
Differentiate the given function with respect to x then,
To apply chain rules of derivatives,
\(\displaystyle{p}'={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({e}^{{{x}^{{{2}}}{\sin{{x}}}}}\right)}\)
\(\displaystyle={e}^{{{x}^{{{2}}}{\sin{{x}}}}}{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{x}^{{{2}}}{\sin{{x}}}\right]}\)
\(\displaystyle={e}^{{{x}^{{{2}}}{\sin{{x}}}}}{\left[{x}^{{{2}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\sin{{x}}}+{\sin{{x}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{x}^{{{2}}}\right]}\)
\(\displaystyle={e}^{{{x}^{{{2}}}{\sin{{x}}}}}{\left[{x}^{{{2}}}{\cos{{x}}}+{\sin{{x}}}\cdot{2}{x}\right]}\)
\(\displaystyle{p}'={e}^{{{x}^{{{2}}}{\sin{{x}}}}}{\left[{x}^{{{2}}}{\cos{{x}}}+{2}{x}{\sin{{x}}}\right]}\)
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