Question #d3912
1 Answer
Mar 3, 2017
Explanation:
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
#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))#
