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differentiationWhat is the differential of y =(A^x -B^x)^(1/x) ? I make it dy/dx = (1/x)*(A^x - B^x)^(1/x-1)*(A^x -B^x)*LOGe(10) = (1/x)*(A^x - B^x)^(1/x)*LOGe(10) but trials show this is wrong, because y change for x change by 1 is less than this increasingly less as x increases. Where am I going wrong and what is the correct differential?:confused::confused::confused: |

See here. (W|A). $$y=(A^x-B^x)^{1/x}$$ Take the natural logarithm of both sides: $$\ln(y)=\frac1x\ln(A^x-B^x)$$ Differentiate implicitly with respect to $x$: $$\frac{y'}{y}=-\frac{1}{x^2}\ln(A^x-B^x)+\frac1x\frac{\ln(A)A^x-\ln(B)B^x}{A^x-B^x}$$ I'll let you finish up. ;) |

ThanksThank you for such a quick reply:) |

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