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magicterry February 24th, 2018 03:57 AM

differentiation
 
What 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:

greg1313 February 24th, 2018 04:04 AM

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. ;)

magicterry February 24th, 2018 04:24 AM

Thanks
 
Thank you for such a quick reply:)


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