Question

$z = log(e^{x} + e^{y}). \: find \: \binom{dz}{dx}$
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Answer 1

Step-by-step explanation:

$z = log( {e}^{x} + {e}^{y} ) \\ \frac{dz}{dx} = \frac{1}{ {e}^{x} + {e }^{y} } \times \frac{d}{dx} ( {e}^{x} + {e}^{y} ) \\ = \frac{1}{ {e}^{x} + {e}^{y} } \times {e}^{x} \\ \frac{dz}{dx} = \frac{ {e}^{x} }{ {e}^{x} + {e}^{y} }$

Any term with y will become constant, whose differentiation with respect to x is 0.