MCQ
If $y = {e^{\sqrt x }}$, then ${{dy} \over {dx}}$ equals
- ✓${{{e^{\sqrt x }}} \over {2\sqrt x }}$
- B${{\sqrt x } \over {{e^{\sqrt x }}}}$
- C${x \over {{e^{\sqrt x }}}}$
- D${{2\sqrt x } \over {{e^{\sqrt x }}}}$
==> $\frac{{dy}}{{dx}} = {e^{\sqrt x }}.\frac{d}{{dx}}\sqrt x $
==> $\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{2\sqrt x }}$.
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$(A)$ $f$ has a local maximum at $x=2$
$(B)$ $f$ is decreasing on $(2,3)$
$(C)$ there exists some $c \in(0, \infty)$ such that $f ^{\prime \prime}( c )=0$
$(D)$ $f$ has a local minimum at $x=3$