the the first principle to find y'(x) of y(x)=(5-3x)/(3-5x)

To find the derivative of the function $y(x) = \frac{5-3x}{3-5x}$, we can use the quotient rule. The quotient rule states that if we have a function $y(x) = \frac{u(x)}{v(x)}$, then its derivative is given by:<br /><br />$y'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$<br /><br />In this case, we have $u(x) = 5-3x$ and $v(x) = 3-5x$. Let's find their derivatives:<br /><br />$u'(x) = -3$<br />$v'(x) = -5$<br /><br />Now, we can apply the quotient rule:<br /><br />$y'(x) = \frac{(-3)(3-5x) - (5-3x)(-5)}{(3-5x)^2}$<br /><br />Simplifying the expression, we get:<br /><br />$y'(x) = \frac{-9 + 15x + 25 - 15x}{(3-5x)^2}$<br />$y'(x) = \frac{16}{(3-5x)^2}$<br /><br />Therefore, the derivative of the function $y(x) = \frac{5-3x}{3-5x}$ is $y'(x) = \frac{16}{(3-5x)^2}$.