Question #d3912

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Answer 1 · 2017-03-03T21:36:40

$dy/dx=-(yx^(y-1)+y^xlny)/(x^ylnx+xy^(x-1))$

Let's tackle each of these parts separately.

$x^y=e^ln(x^y)=e^(ylnx)$

Then:

$d/dxx^y=d/dxe^(ylnx)=e^(ylnx)d/dx(ylnx)$

Which can be found through the product rule:

$d/dxx^y=e^(ylnx)(dy/dxlnx+y/x)=x^y(dy/dxlnx+y/x)$

The other:

$y^x=e^ln(y^x)=e^(xlny)$

So:

$d/dxy^x=d/dxe^(xlny)=e^(xlny)d/dx(xlny)$

Product rule:

$d/dxy^x=e^(xlny)(lny+x/ydy/dx)=y^x(lny+x/ydy/dx)$

So, we see that the differentiated implicit function becomes:

$x^y+y^x=1$

$=>x^y(dy/dxlnx+y/x)+y^x(lny+x/ydy/dx)=0$

Expanding and keeping all terms with $dy/dx$ on one side:

$x^ylnxdy/dx+y^x x/ydy/dx=-x^yy/x-y^xlny$

Factoring and simplifying exponents:

$dy/dx(x^ylnx+xy^(x-1))=-(yx^(y-1)+y^xlny)$

$dy/dx=-(yx^(y-1)+y^xlny)/(x^ylnx+xy^(x-1))$