Basic maths-calculus-Q9

+1 vote

x^y = y^x find   \frac{dy}{dx}

(A)  \frac{x(ylogx - y)}{y(xlogx-x)}

(B)  \frac{y(xlogy - y)}{x(ylogx-x)}

(C)  \frac{y(ylogx - x)}{x(ylogx-x)}

(D)  \frac{x(xlogy - y)}{x(ylogx-x)}

asked May 19 in Basic Maths by gbeditor (44,490 points)
reshown May 24 by gbeditor

1 Answer

0 votes
 
Best answer

Taking log on both sides:::

log(x^y)=log(y^x)\\ \\ ylogx=xlogy\\ \text{differentiating\ both\ sides\ wrt\ to\ x} \\ \\ y*1/x+logx\ *dy/dx=x/y*dy/dx+logy\\ \\ logx*dy/dx-x/y*dy/dx=logy-y/x\\ \\ dy/dx[logx-x/y]=logy-y/x\\ \\ dy/dx[(ylogx-x)/y]=(xlogy-y)/x\\ \\ dy/dx=y*(xlogy-y)/x(ylogx-x)\\ \\

 

Therefore Answer should be B.

answered May 27 by deepak-gatebook (112,390 points)
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