I know that the answer is dy/dx = 6^x In x. I just do not understand the steps!
Thanks for ANY help!!!
2007-12-06 05:31:06 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
log y = x log 6
d/dx (log y) = log 6
d/dy(log y) (dy/dx) = log 6
(1/y) (dy/dx) = log 6
(dy/dx) = y log 6
dy/dx = (log 6)(6^x)
2007-12-06 05:48:56 · answer #1 · answered by Como 7 · 2⤊ 0⤋
y= 6ⁿ
y'=?
well lets just take this as our equation for a sec:
y= kⁿ where k = a constant
y' = ln(k) kⁿ
so y'= 6ⁿ =
y' = ln(6) 6ⁿ
Just a rule you have to remember.
It may look like the eⁿ which when you take the derivative is itself but that is a different case so be careful not to get those two mixed up.
Hope that was helpful!
2007-12-06 05:36:30 · answer #2 · answered by Matty B 3 · 0⤊ 0⤋
y=6^x
Take ln on both sides
ln y = x ln (6)
Differentiate both sides
(1/y) dy/dx = ln (6)
Multiply both sides by y or solve for dy/dx
dy/dx=y ln(6)
i.e. dy/dx=6^x ln(6)
2007-12-06 05:41:53 · answer #3 · answered by cidyah 7 · 0⤊ 0⤋
y = 6^x
take natural log on both sides
ln(y) = ln(6^x)
ln(y) = xln(6) [since lna^b = b*ln(a)]
now differentiate with respect to x using product rule
(1/y)dy/dx = x*(0) + ln(6)*1
(1/y)dy/dx = ln(6)
multiply with y
dy/dx = y*ln(6)
substitute back y = 6^x
dy/dx = 6^x ln(6)
2007-12-06 05:50:57 · answer #4 · answered by mohanrao d 7 · 0⤊ 1⤋
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2007-12-06 05:37:07 · answer #5 · answered by Vaneet P 1 · 0⤊ 0⤋