Derivative of $$$e^{\sin{\left(x \right)}}$$$
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Solution
The function $$$e^{\sin{\left(x \right)}}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = e^{u}$$$ and $$$g{\left(x \right)} = \sin{\left(x \right)}$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(e^{\sin{\left(x \right)}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}$$The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$Return to the old variable:
$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = e^{{\color{red}\left(\sin{\left(x \right)}\right)}} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$The derivative of the sine is $$$\frac{d}{dx} \left(\sin{\left(x \right)}\right) = \cos{\left(x \right)}$$$:
$$e^{\sin{\left(x \right)}} {\color{red}\left(\frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)} = e^{\sin{\left(x \right)}} {\color{red}\left(\cos{\left(x \right)}\right)}$$Thus, $$$\frac{d}{dx} \left(e^{\sin{\left(x \right)}}\right) = e^{\sin{\left(x \right)}} \cos{\left(x \right)}$$$.
Answer
$$$\frac{d}{dx} \left(e^{\sin{\left(x \right)}}\right) = e^{\sin{\left(x \right)}} \cos{\left(x \right)}$$$A