In a specific chemical titration, the potential difference VVV across an electrode is modeled by the equation V=ln(0.004t)V = \ln(0.004t)V=ln(0.004t), where t>0t > 0t>0 is the time in seconds since the reaction began.
Determine an expression for the rate of change of the potential difference with respect to time, dVdt\frac{dV}{dt}dtdV.
Select the correct option:
A: dVdt=1t\frac{dV}{dt} = \frac{1}{t}dtdV=t1
B: dVdt=0.004t\frac{dV}{dt} = \frac{0.004}{t}dtdV=t0.004
C: dVdt=10.004t\frac{dV}{dt} = \frac{1}{0.004t}dtdV=0.004t1
D: dVdt=ln(0.004)\frac{dV}{dt} = \ln(0.004)dtdV=ln(0.004)
375 exam-style questions on AQA A Level Maths 1.10 G: Differentiation, covering 1.10.1 The derivative and second derivative, 1.10.2 Differentiating standard functions, 1.10.3 Applications of differentiation, 1.10.4 Product, quotient and chain rules (A-level only), 1.10.5 Implicit and parametric differentiation (A-level only), and 1.10.6 Constructing differential equations (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.