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1.9.10 Differentiate e^kx, a^kx and ln x (A-level only)

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Question 8

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)

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1.9.10 Differentiate e^kx, a^kx and ln x (A-level only) Questions

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