y = 4x - 1/x defines y as f of x explicitly xy - 4x + 1 = 0 defines y as f of x
ID: 2881554 • Letter: Y
Question
y = 4x - 1/x defines y as f of x explicitly xy - 4x + 1 = 0 defines y as f of x implicitly x^2 + y^2 - 9 = 0 y = plusminus Squareroot 9 - x^2 y is not a f of x, but we can consider separately y = Squareroot 9 - x^2 and y = - Squareroot 9 - x^2 so x^2 + y^2 - 9 = 0 implicitly defines 2 functions. We are implying that y is a f of x without verifying it. We dy/dx. Simply derivative of both sides of solve algebraically for dy/dx. d/dx [x^2 + y^2 - 9] = d/dx (0). 2x + d/dx y^2 + 0 = 0 2x + 2y dy/dx = 0 2y dy/dx = -2x dy/dx = -2x/2y = -x/y Since we assumed y is a f of x the derivative of y^2 is treated like derivative of u^n where u is f of x.Explanation / Answer
2) we have given xy-4x+1=0
differentiating with respect to x
xdy/dx+y-4=0
dy/dx=(-y+4)/x
dy/dx =-y/x+4/x
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