Question

If $v(z) = -11e^{2z} \ln(z^3)$, then: a) $v'(z) = -11e^{2z} \ln(z^3) - \frac{66e^{2z}}{z}$ b) $v'(z) = -11e^{2z} \ln(z^3) - 22e^{2z}z \ln(z^3) - 33e^{2z}$ c) $v'(z) = -\frac{66e^{2z}}{z}$ d) $v'(z) = -22e^{2z} \ln(z^3) - \frac{33e^{2z}}{z}$

          If $v(z) = -11e^{2z} \ln(z^3)$, then:
a) $v'(z) = -11e^{2z} \ln(z^3) - \frac{66e^{2z}}{z}$
b) $v'(z) = -11e^{2z} \ln(z^3) - 22e^{2z}z \ln(z^3) - 33e^{2z}$
c) $v'(z) = -\frac{66e^{2z}}{z}$
d) $v'(z) = -22e^{2z} \ln(z^3) - \frac{33e^{2z}}{z}$

If v(z) = -11e^2zln(z^3), then:
a) v'(z) = -11e^2zln(z^3) - (66e^2z)/(z)
b) v'(z) = -11e^2zln(z^3) - 22e^2zz ln(z^3) - 33e^2z
c) v'(z) = -(66e^2z)/(z)
d) v'(z) = -22e^2zln(z^3) - (33e^2z)/(z)

Added by Francisco Javier W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/25/2024

Step 1

Step 1: Use the product rule to differentiate $v(z)$. Show more…

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If $v(z) = -11e^{2z}ln(z^3)$, then: a) $v'(z) = -11e^{2z}ln(z^3) - \frac{66e^{2z}}{z}$ b) $v'(z) = -11e^{2z}ln(z^3) - 22e^{2z}zln(z^3) - 33e^{2z}$ c) $v'(z) = -\frac{66e^{2z}}{z}$ d) $v'(z) = -22e^{2z}ln(z^3) - \frac{33e^{2z}}{z}$