Derivative of $$$e^{x} + \sin{\left(y z \right)}$$$ with respect to $$$x$$$
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Your Input
Find $$$\frac{d}{dx} \left(e^{x} + \sin{\left(y z \right)}\right)$$$.
Solution
The derivative of a sum/difference is the sum/difference of derivatives:
$${\color{red}\left(\frac{d}{dx} \left(e^{x} + \sin{\left(y z \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(e^{x}\right) + \frac{d}{dx} \left(\sin{\left(y z \right)}\right)\right)}$$The derivative of the exponential is $$$\frac{d}{dx} \left(e^{x}\right) = e^{x}$$$:
$${\color{red}\left(\frac{d}{dx} \left(e^{x}\right)\right)} + \frac{d}{dx} \left(\sin{\left(y z \right)}\right) = {\color{red}\left(e^{x}\right)} + \frac{d}{dx} \left(\sin{\left(y z \right)}\right)$$The derivative of a constant is $$$0$$$:
$$e^{x} + {\color{red}\left(\frac{d}{dx} \left(\sin{\left(y z \right)}\right)\right)} = e^{x} + {\color{red}\left(0\right)}$$Thus, $$$\frac{d}{dx} \left(e^{x} + \sin{\left(y z \right)}\right) = e^{x}$$$.
Answer
$$$\frac{d}{dx} \left(e^{x} + \sin{\left(y z \right)}\right) = e^{x}$$$A