y=(4x^3)/(2x^2)-5 Find the derivative of y (8x^4-60x^2)/(2x^2-5)^2 16x^4 3× 8x^4-60x^2

To find the derivative of the function \( y = \frac{4x^3}{2x^2 - 5} \), we will use the quotient rule. The quotient rule states that if you have a function \( y = \frac{u(x)}{v(x)} \), then its derivative is given by:<br /><br />\[ y' = \frac{u'v - uv'}{v^2} \]<br /><br />Here, \( u(x) = 4x^3 \) and \( v(x) = 2x^2 - 5 \).<br /><br />First, we find the derivatives of \( u(x) \) and \( v(x) \):<br /><br />\[ u'(x) = 12x^2 \]<br />\[ v'(x) = 4x \]<br /><br />Now, apply the quotient rule:<br /><br />\[ y' = \frac{(12x^2)(2x^2 - 5) - (4x^3)(4x)}{(2x^2 - 5)^2} \]<br /><br />Simplify the numerator:<br /><br />\[ y' = \frac{24x^4 - 60x^2 - 16x^4}{(2x^2 - 5)^2} \]<br />\[ y' = \frac{8x^4 - 60x^2}{(2x^2 - 5)^2} \]<br /><br />So, the derivative of \( y = \frac{4x^3}{2x^2 - 5} \) is:<br /><br />\[ y' = \frac{8x^4 - 60x^2}{(2x^2 - 5)^2} \]