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Q. If $y = 2^{-x}$, then $\frac{dy}{dx}$ is equal to :

Continuity and Differentiability

Solution:

Let $y = 2^{-x}$
$\frac{dy}{dx} =\frac{2^{-x}}{\log 2} \left(-1\right)=\frac{-1}{2^{x} \log 2} \frac{\log \frac{1}{2}}{2^{x}}$