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

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

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Sue Leahy
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,
$$\displaystyle{p}'={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({e}^{{{x}^{{{2}}}{\sin{{x}}}}}\right)}$$
$$\displaystyle={e}^{{{x}^{{{2}}}{\sin{{x}}}}}{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{x}^{{{2}}}{\sin{{x}}}\right]}$$
$$\displaystyle={e}^{{{x}^{{{2}}}{\sin{{x}}}}}{\left[{x}^{{{2}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\sin{{x}}}+{\sin{{x}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{x}^{{{2}}}\right]}$$
$$\displaystyle={e}^{{{x}^{{{2}}}{\sin{{x}}}}}{\left[{x}^{{{2}}}{\cos{{x}}}+{\sin{{x}}}\cdot{2}{x}\right]}$$
$$\displaystyle{p}'={e}^{{{x}^{{{2}}}{\sin{{x}}}}}{\left[{x}^{{{2}}}{\cos{{x}}}+{2}{x}{\sin{{x}}}\right]}$$