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)
Practise Edexcel A Level Maths Differentiation with exam-style questions for A Level Maths. 311 questions 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, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
