Question #d3912

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Question #d3912

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Answer 1 · 2017-03-03T21:36:40

$\frac{d y}{d x} = - \frac{y x^{y - 1} + y^{x} ln y}{x^{y} ln x + x y^{x - 1}}$ $dy/dx=-(yx^(y-1)+y^xlny)/(x^ylnx+xy^(x-1))$

Explanation:

Let's tackle each of these parts separately.

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

Then:

$\frac{d}{d x} x^{y} = \frac{d}{d x} e^{y ln x} = e^{y ln x} \frac{d}{d x} (y ln x)$ $d/dxx^y=d/dxe^(ylnx)=e^(ylnx)d/dx(ylnx)$

Which can be found through the product rule:

$\frac{d}{d x} x^{y} = e^{y ln x} (\frac{d y}{d x} ln x + \frac{y}{x}) = x^{y} (\frac{d y}{d x} ln x + \frac{y}{x})$ $d/dxx^y=e^(ylnx)(dy/dxlnx+y/x)=x^y(dy/dxlnx+y/x)$

The other:

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

So:

$\frac{d}{d x} y^{x} = \frac{d}{d x} e^{x ln y} = e^{x ln y} \frac{d}{d x} (x ln y)$ $d/dxy^x=d/dxe^(xlny)=e^(xlny)d/dx(xlny)$

Product rule:

$\frac{d}{d x} y^{x} = e^{x ln y} (ln y + \frac{x}{y} \frac{d y}{d x}) = y^{x} (ln y + \frac{x}{y} \frac{d y}{d x})$ $d/dxy^x=e^(xlny)(lny+x/ydy/dx)=y^x(lny+x/ydy/dx)$

So, we see that the differentiated implicit function becomes:

$x^{y} + y^{x} = 1$ $x^y+y^x=1$

$\Rightarrow x^{y} (\frac{d y}{d x} ln x + \frac{y}{x}) + y^{x} (ln y + \frac{x}{y} \frac{d y}{d x}) = 0$ $=>x^y(dy/dxlnx+y/x)+y^x(lny+x/ydy/dx)=0$

Expanding and keeping all terms with $\frac{d y}{d x}$ $dy/dx$ on one side:

$x^{y} ln x \frac{d y}{d x} + y^{x} \frac{x}{y} \frac{d y}{d x} = - x^{y} \frac{y}{x} - y^{x} ln y$ $x^ylnxdy/dx+y^x x/ydy/dx=-x^yy/x-y^xlny$

Factoring and simplifying exponents:

$\frac{d y}{d x} (x^{y} ln x + x y^{x - 1}) = - (y x^{y - 1} + y^{x} ln y)$ $dy/dx(x^ylnx+xy^(x-1))=-(yx^(y-1)+y^xlny)$

$\frac{d y}{d x} = - \frac{y x^{y - 1} + y^{x} ln y}{x^{y} ln x + x y^{x - 1}}$ $dy/dx=-(yx^(y-1)+y^xlny)/(x^ylnx+xy^(x-1))$

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Question #d3912

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