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)
425 exam-style questions on OCR A Level Maths 1.7 Differentiation, covering 1.7.1 Derivative as gradient of the tangent, 1.7.2 Gradient of the tangent at a point, 1.7.3 Sketching the gradient function, 1.7.4 Second derivatives, 1.7.5 Second derivative as rate of change of gradient, 1.7.6 Convex, concave and points of inflection (A-level only), 1.7.7 Differentiation from first principles for powers of x, 1.7.8 Differentiation from first principles for sin x and cos x (A-level only), 1.7.9 Differentiating x^n, 1.7.10 Differentiating e^(kx) and a^(kx) (A-level only), 1.7.11 Differentiating trigonometric functions (A-level only), 1.7.12 Derivative of ln x (A-level only), 1.7.13 Tangents and normals, 1.7.14 Stationary points, 1.7.15 Increasing and decreasing functions, 1.7.16 Points of inflection (A-level only), 1.7.17 Product and quotient rules (A-level only), 1.7.18 Chain rule (A-level only), 1.7.19 Parametric and implicit differentiation (A-level only), 1.7.20 Constructing differential equations (A-level only), and 1.7 Differentiation. Each one has a worked solution and a mark scheme showing where the marks go.