(c) $E = \frac{1}{2}mv^{2}$. Differentiating $w.r.t.$ $x$, we get $\frac{dE}{dx} = \frac{1}{2}m \times 2v\frac{dv}{dx} = mv \times \frac{dv}{dt} \times \frac{dt}{dx} = mv \times \frac{a}{v} = ma$
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Ph.D. in Physics with 19+ years of teaching experience.