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)
717 exam-style questions on Edexcel A Level Maths Differentiation, covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change. Each one has a worked solution and a mark scheme showing where the marks go.