Question

find dy/dx of y=x^7(3^x)

Answer 1

\[(fg)'=f'g+fg'\]

Answer 2

dy/dx = f'(x)g(x) + f(x)g(x)
= (x^7)' * (3^x) + (x^7) * (3^x)'
= [7x^6 * 3^x] + [x^7 * ln(3) * 3^x]
= 3^x * [7x^6 + [x^7 *ln(3)]]

Answer 3

good but \[(3^x)'=x \ln 3*3^x\]

Answer 4

Hmmm. x*ln(3) would imply there that the natural log was ln(3^x), but that's not the case if you look at the original definition:

d/dx (a^x) = ln(a) * a^x

Check it out: http://www.math.com/tables/derivatives/more/b%5Ex.htm

Answer 5

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Answer 6

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Answer 7

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Answer 8

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Answer 9

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Answer 10

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Answer 11

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Answer 12

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