Question 998820: Hi, I need to find the first and second derivative of: f(x)=8x[4^(-x)]
and the answer in the book is:
f'(x)=-8[4^(-x)](x ln 4-1)
and the second one is:
f''(x)=8[4^(-x)] ln 4(x ln 4-2)
can someone show me the steps to get to those derivatives please?
Thank you!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! f(x) = 8x[4^(-x)] = 8x / (4^x)
note that the calculus division rule will be used, that is
first derivative of (numerator(N) / denominator(D)) =
(N' * D)-(N * D') / (D^2)
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derivative of denominator calculation
let y = 4^x
take natural log of both sides of =
ln y = x * ln(4)
then use implicit differentiation
(1/y) * y' = ln(4)
y' = y * ln(4)
note that y = 4^x, then
y' = (4^x) * ln(4)
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derivative of numerator calculation
y = 8x
y' = 8
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f'(x) = (8 * (4^x)) - (8x * (4^x) * ln(4) ) / (4^2x)
simplify this (see the following steps)
f'(x) = (8 - 8x * ln(4)) / (4^x)
here we have divided the numerator by (4^x) leaving (4^x) in the D
f'(x) = (-8 * (x*ln(4)-1) / (4^x)
factored -8, therefore
f'(x) = -8 * [4^(-x)] * (x*ln(4)-1)
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start with f'(x) = (8 - 8x * ln(4)) / (4^x), then take first derivative of this using calculus division rule and implicit derivative of D
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derivative of denominator calculation
let y = 4^x
y' = (4^x) * ln(4)
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derivative of numerator calculation
y = 8 - 8x * ln(4)
note that derivative of a constant is 0
y' = 0 -8*ln(4)
y' = -8*ln(4)
*****************************************************************************
f''(x) = (-(8*ln(4))*(4^x)) - ((8 - 8x * ln(4))*(4^x) * ln(4)) / (4^2x)
f''(x) = -8*ln(4)*4^x -8*ln(4)*4^x +8x*ln^2(4)*4^x / 4^2x
f''(x) = -8*ln(4) -8*ln(4) + 8x*ln^2(4) / 4^x
f''(x) = 8x*ln^2(4) -16*ln(4) / 4^x
f''(x) = 8 * [4^(-x)] * ln(4) * (x*ln(4) - 2)
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