by implicit differentiation you have
y+x*dy/dx-2dy/dx=0
solving for dy/dx you'll have
dy/dx=y/(2-x)
and solving for y in the original equation and plugging it back in, you'll get
dy/dx=1/(-x^2 +4x-4) which is your final answer
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To find the derivative of the equation xy - 2y = 1, differentiate each term with respect to x using the product rule for the first term and the power rule for the second term. This yields dy/dx = (1 - 2x).
f(x)=1 f'(x)=0 because the derivative of a constant is ALWAYS 0.
(xlnx)' = lnx + 1
Yes, the derivative of xi with respect to x equals i. Is that what you were trying to ask?
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
The derivative of 10x is 10. This is irrespective of the value of x.