Using the quotient rule, \( \frac{d y}{d x}=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}} \) the derivative of \( y=\frac{x^{5}+5 x^{3}+5 x^{2}}{3 x^{2}+3} \) is \( \frac{d y}{d x}=\frac{\left(5 x^{4}+15 x^{2}+10 x\right)\left(3 x^{2}+3\right)-6 x\left(x^{5}+5 x^{3}+5 x^{2}\right)}{\left(3 x^{2}+3\right)^{2}} \) Is the derivative TRUE or FALSE?

Let's break down the problem to see if the derivative is indeed correct! The function given is \( y = \frac{u}{v} \), where \( u = x^5 + 5x^3 + 5x^2 \) and \( v = 3x^2 + 3 \). Following the quotient rule, we compute the derivatives of \( u \) and \( v \): The derivative of \( u \) is \( \frac{du}{dx} = 5x^4 + 15x^2 + 10x \), and the derivative of \( v \) is \( \frac{dv}{dx} = 6x \). Plugging these into the quotient rule, we get: \[ \frac{dy}{dx} = \frac{(v)(\frac{du}{dx}) - (u)(\frac{dv}{dx})}{v^2} \] Substituting back, we find \[ \frac{dy}{dx} = \frac{(3x^2 + 3)(5x^4 + 15x^2 + 10x) - (x^5 + 5x^3 + 5x^2)(6x)}{(3x^2 + 3)^2} \] Now, comparing this with the expression you provided, it matches! Therefore, the derivative is TRUE. Now, let's take a look at how this ties into real-world applications and some common problems you might face while dealing with derivatives. In the real world, the quotient rule is used extensively in fields like physics and economics. For instance, when calculating rates of change where one quantity depends on another that also varies, such as speed (distance/time) or profit margins (profit/revenue), applying the quotient rule helps ascertain how these varying relationships change — think of it as decoding the underlying patterns in data! When applying the quotient rule, a common mistake is forgetting to square the denominator after differentiation. It’s like going to a party but leaving behind the gift you promised! Always double-check your final denominator to ensure completeness in your calculations.