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Find the Derivative, Using Chain Rule. y=2x^2-4x+3/2-3x
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\[y=\frac{ 2x^2-4x+3 }{ 2-3x }\]
You can start out by using the quotient rule for this problem
\[\frac{ (2-3x)d/dx[2x^2-4x+3)-d/dx(2-3x)(2x^2-4x+3)] }{ (2-3x)^2 }\]
So the chain rule wouldn't apply?
The equation should have an inner and an outer layer in order for us to take the derivative of the outer and then multiply it by the inner's derivative.
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maybe they want you to think of it as y = (2x^2-4x+3)*(2-3x)^(-1) then use the product rule
you'd use the chain rule when it comes to differentiating (2-3x)^(-1)
whyd you bring the demominator up?
if you want to multiply the denominator by the numerator, you take the reciprocal
for example |dw:1414371954419:dw| 1/x is the same as x^-1
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