Let h(x)=(x^(2)+1)/(2x^(2)-3x). h^(')(x)=

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Let  h(x)=(x^(2)+1)/(2x^(2)-3x).  h^(')(x)=

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Apply Quotient Rule: To find the derivative of the function h(x), we will use the quotient rule, which states that the derivative of a function that is the quotient of two functions, u(x)/v(x), is given by (v(x)u'(x) - u(x)v'(x)) / (v(x))^2. Here, u(x) = x^2 + 1 and v(x) = 2x^2 - 3x.Find u'(x): First, we need to find the derivative of u(x), which is u'(x). The derivative of x^2 is 2x, and the derivative of a constant is 0. Therefore, u'(x) = 2x + 0 = 2x.Find v'(x): Next, we need to find the derivative of v(x), which is v'(x). The derivative of 2x^2 is 4x, and the derivative of -3x is -3. Therefore, v'(x) = 4x - 3.Simplify Numerator: Now we apply the quotient rule. We have: h'(x) = ( (2x^2 - 3x)(2x) - (x^2 + 1)(4x - 3) ) / (2x^2 - 3x)^2.Subtract Terms: We simplify the numerator of the derivative: (2x^2 - 3x)(2x) = 4x^3 - 6x^2, (x^2 + 1)(4x - 3) = 4x^3 - 3x^2 + 4x - 3. So, h'(x) = (4x^3 - 6x^2 - (4x^3 - 3x^2 + 4x - 3)) / (2x^2 - 3x)^2.Final Derivative: Subtract the terms in the numerator: h'(x) = (4x^3 - 6x^2 - 4x^3 + 3x^2 - 4x + 3) / (2x^2 - 3x)^2 = (-3x^2 + 3x^2 - 4x + 3) / (2x^2 - 3x)^2 = (-4x + 3) / (2x^2 - 3x)^2.We have found the derivative of h(x) in its simplified form: h'(x) = (-4x + 3) / (2x^2 - 3x)^2.

Let $h(x)=\frac{x^{2}+1}{2 x^{2}-3 x}$ . $\newline$ $h^{\prime}(x)=$

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Let $h(x)=\frac{x^{2}+1}{2 x^{2}-3 x}$ . $\newline$ $h^{\prime}(x)=$