Just as there are times when the erroneous version of the product rule holds (see the previous exercise), there are times when the erroneous version of the quotient rule holds. Calcu- late [f(x)/g(x)]' and f'(x)/g'(x) for f(x) = ae*k-1) and g(x) = be* for k + 1, and verify that they are equal. Source: PRIMUS.
Just as there are times when the erroneous version of the product rule holds (see the previous exercise), there are times when the erroneous version of the quotient rule holds. Calcu- late [f(x)/g(x)]' and f'(x)/g'(x) for f(x) = ae*k-1) and g(x) = be* for k + 1, and verify that they are equal. Source: PRIMUS.
Chapter6: Exponential And Logarithmic Functions
Section6.4: Graphs Of Logarithmic Functions
Problem 60SE: Prove the conjecture made in the previous exercise.
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![Just as there are times when the erroneous version of the
product rule holds (see the previous exercise), there are times
when the erroneous version of the quotient rule holds. Calcu-
late [f(x)/g(x)]' and f'(x)/g'(x) for f(x) = ae*k-1) and
g(x) = be* for k + 1, and verify that they are equal. Source:
PRIMUS.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07c94bfc-ec0f-4a99-b286-1960ad911e6e%2F87a65b99-8c51-45f2-a8e8-64f008b78f3c%2F2gdmbuo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Just as there are times when the erroneous version of the
product rule holds (see the previous exercise), there are times
when the erroneous version of the quotient rule holds. Calcu-
late [f(x)/g(x)]' and f'(x)/g'(x) for f(x) = ae*k-1) and
g(x) = be* for k + 1, and verify that they are equal. Source:
PRIMUS.
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