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What is the linear approximation to ln((1+𝑎𝑥)^𝑟) at 𝑥=0 ?
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Answered by
oobleck
If y = ln(1+ax)^r = r ln(1+ax)
y' = ra/(1+ax)
y(0) = 0
y'(0) = ra
so, the tangent line at (0,0) is
y = rax
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