Detail solution
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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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Differentiate term by term:
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The derivative of cosine is negative sine:
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The derivative of is .
The result is:
The result of the chain rule is:
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Now simplify:
The answer is:
The first derivative
[src]
cos(t) /1 \
t*e *|- - sin(t)|
\t /
$$t e^{\cos{\left(t \right)}} \left(- \sin{\left(t \right)} + \frac{1}{t}\right)$$
The second derivative
[src]
/ / 1 \ /1 \\ cos(t)
|(-1 + t*sin(t))*|- - + sin(t)| - t*|-- + cos(t)||*e
| \ t / | 2 ||
\ \t //
$$\left(- t \left(\cos{\left(t \right)} + \frac{1}{t^{2}}\right) + \left(t \sin{\left(t \right)} - 1\right) \left(\sin{\left(t \right)} - \frac{1}{t}\right)\right) e^{\cos{\left(t \right)}}$$
The third derivative
[src]
/ 1 /2 \ /1 \ / 1 \ / 2 \ /1 \ \ cos(t)
|- -- - cos(t) + t*|-- + sin(t)| + (-1 + t*sin(t))*|-- + cos(t)| + |- - + sin(t)|*\2*sin(t) + t*cos(t) - t*sin (t)/ + t*|-- + cos(t)|*sin(t)|*e
| 2 | 3 | | 2 | \ t / | 2 | |
\ t \t / \t / \t / /
$$\left(t \left(\cos{\left(t \right)} + \frac{1}{t^{2}}\right) \sin{\left(t \right)} + t \left(\sin{\left(t \right)} + \frac{2}{t^{3}}\right) + \left(t \sin{\left(t \right)} - 1\right) \left(\cos{\left(t \right)} + \frac{1}{t^{2}}\right) + \left(\sin{\left(t \right)} - \frac{1}{t}\right) \left(- t \sin^{2}{\left(t \right)} + t \cos{\left(t \right)} + 2 \sin{\left(t \right)}\right) - \cos{\left(t \right)} - \frac{1}{t^{2}}\right) e^{\cos{\left(t \right)}}$$