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logarithmic differentiation.

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Differentiate y = (9+x)^x with respect to x using logarithmic differentiation.

 

asked Jul 6, 2013 in ALGEBRA 2 by harvy0496 Apprentice

1 Answer

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Given that y  = (9 + x)^x

Taking logarithms on both sides we get,

ln(y) = ln(9+x)^x

ln(y) = xln(9+x)                                                      [Since ln(a)^n = nln(a)]

Differentiating on both sides we get,

d/dx(ln(y)) = d/dx( xln(9+x) )

dy/dx(1/y) = xd/dx( ln(9+x) ) + ln(9+x)d/dx(x)         [Since d/dx(uv) = udv/dx + vdu/dx]

dy/dx(1/y) = ln(9+x) + x/(9+x)                                 [Since d/dx(x) = 1, d/dx(logx) = 1/x]

dy/dx = y* [ln(9+x) + x/(9+x)]

dy/dx = (9+x)^x * [ln(9+x)+x/(9+x)]                          [Since y = (9 + x)^x]

Therefore dy/dx = (9+x)^x * [ln(9+x)+x/(9+x)]

answered Jul 6, 2013 by joly Scholar

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