Detail solution
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Apply the product rule:
; to find :
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Let .
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The derivative of is itself.
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Then, apply the chain rule. Multiply by :
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The derivative of sine is cosine:
The result of the chain rule is:
; to find :
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The derivative of sine is cosine:
The result is:
Now simplify:
The answer is:
The first derivative
[src]
sin(x) sin(x)
cos(x)*e + cos(x)*e *sin(x)
$$e^{\sin{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)} + e^{\sin{\left(x \right)}} \cos{\left(x \right)}$$
The second derivative
[src]
/ 2 / 2 \ \ sin(x)
\-sin(x) + 2*cos (x) - \- cos (x) + sin(x)/*sin(x)/*e
$$\left(- \left(\sin{\left(x \right)} - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} - \sin{\left(x \right)} + 2 \cos^{2}{\left(x \right)}\right) e^{\sin{\left(x \right)}}$$
The third derivative
[src]
/ 2 / 2 \ \ sin(x)
-\1 - 3*cos (x) + 6*sin(x) + \1 - cos (x) + 3*sin(x)/*sin(x)/*cos(x)*e
$$- \left(\left(3 \sin{\left(x \right)} - \cos^{2}{\left(x \right)} + 1\right) \sin{\left(x \right)} + 6 \sin{\left(x \right)} - 3 \cos^{2}{\left(x \right)} + 1\right) e^{\sin{\left(x \right)}} \cos{\left(x \right)}$$