Find the nth differential coefficient of log(1+x).
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This is how I would do it
f(x)=ln(1+x)
f′(x)=1x+1
f′′(x)=−1(x+1)2
f′′′(x)=(−1)(−2)1(x+1)3
f′v(x)=(−1)(−2)(−3)1(x+1)4
f′v(x)=(−1)3⋅3!(x+1)4
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f′n(x)=(−1)n−1⋅(n−1)!(x+1)n∀n≥1
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