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Find the derivative of each function by using the quotient rule.$$y=\frac{e^{2}}{x(3 x-5)}$$

$\frac{d y}{d x}=\frac{5 e^{2}-6 e^{2} x}{\left(3 x^{2}-5 x\right)^{2}}$

Calculus 1 / AB

Chapter 23

The Derivative

Section 6

Derivatives of Products and Quotients of Functions

Derivatives

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All right, We are given why equals e squared all over X times three x minus five. Uh Now we are going to do the quotient rule, There's a quotient in there um but before I get started, just remember that E squared is a constant. And why am I bringing that up is because the derivative of a constant is zero. So when I go to take the derivative using the question rule, Um the derivative of the top is zero. That's what I'm trying to get at. Um And you leave the bottom alone and then minus. Now you leave the tap alone. But notice as we do the drift of the bottom, we have the product rule. So you have to do The quantity of the derivative of X, which would be one. You leave three X -5 alone. And then plus, that's why I put in parentheses, we have to distribute. I guess I should have put parentheses right here. Yeah, because you got multiplied by both pieces, um then leave X alone and then times the derivative of three x minus five, which would just be three all over the denominator squared, just square each piece you'll be fine. So as you notice this zero times anything will be zero, so that's gone. Um I can actually simplify inside here that I have three X plus three X six X uh minus five. Uh So let's actually leave the problem like that because nothing fancy is going to happen. So I'm gonna write my answer like this negative of E squared Times The quantity of six x -5 all over that X squared three x minus five squared because again nothing canceled. So this should be good answer for you.

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