Find the derivative of the function. y = 2 − 3 t 4 t 2 + 7 y=\frac{2-3t}{4t^2+7}

In this problem, we're asked to find the derivative of the function y equals 2 minus 3t over 4t squared plus 7. So what can we do here to find this derivative? Well, it's clear that there are 2 functions being divided. So that means I need to use the quotient rule in order to find this derivative. Now to find my derivative, I can either denote this as y prime or I can denote this as a dydt. Either of these are correct. Now let's get started in applying this quotient rule. Remember our quotient rule tells us that we're going to take that high function and that low function, and we remember our quotient rule by saying low d high minus high d low over the square of what's below. So applying that quotient rule here, I'm going to take my low function, that 4 t squared plus 7, starting with that, 4 t squared plus 7 times the derivative of that top function. Now the derivative of this high function, that top function here, is just going to be a negative 3 because the derivative of 2 is just 0, and then the derivative of negative 3 t is negative 3. So that's my first term in my numerator here. I am subtracting. So I have a low d high minus high d low. So taking that top function, 2 minus 3 t times the derivative of that bottom function, using my power rule here, if I pull that 2 out to the front, decrease that power by 1, that is going to give me 8 t to the power of 2 minus 1 which is just 1. So this is my full numerator. Now all of this is being divided by the square of what's below that's 4 t squared plus 7 and all of that is squared over the square of what's below. Don't forget that square. Now from here we can do some simplification. We have successfully applied our quotient rule and it's all algebra from here which is often the most tedious part so bear with me here. Now we're going to pull this negative 3 into these parentheses distributing it. We're going to do the same thing with this 8 t and doing that we end up with a numerator of negative 12 t squared minus 21 and then subtracting here I'm going to keep all that parentheses for now 8t times 2 is 16t and then a minus 24t squared. Now that denominator stays as is 4t squared plus plus 7 squared. Now from here I'm going to distribute this negative in here. Remember that we are subtracting so we want to keep this organized here. Negative 12t squared minus 21, haven't changed anything there. Then I have minus 16t plus 24 t squared distributing that negative to that negative giving me a positive. Now that denominator is still the same 4 t squared plus 7 squared. Now I can do a little bit more simplification here because I have some like terms. I have negative 12 t squared and positive 24 t squared. So that's gonna make my numerator here positive 12 t squared. And I'm gonna reorder this a little bit just writing it in standard form. I have minus 16 t minus 21 and then all of that is divided by that same denominator. Make sure you're rewriting it correctly here 4t squared+7 squared, and this is our final answer for our derivative of our function dydt, and we are all done here. Let me know if you have any questions, and I'll see you in the next one.

3. Techniques of Differentiation

Product and Quotient Rules

Calculus

3. Techniques of Differentiation

Product and Quotient Rules

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Multiple Choice

Find the derivative of the function.
$y = \frac{2 - 3 t}{4 t^{2} + 7}$

$\frac{12 t^{2} - 16 t - 21}{(4 t^{2} + 7)^{2}}$

$\frac{- 12 t^{2} + 16 t + 21}{(4 t^{2} + 7)^{2}}$

$\frac{- 12 t^{2} + 16 t + 21}{(2 - 3 t)^{2}}$

$\frac{12 t^{2} - 16 t - 21}{(2 - 3 t)^{2}}$

Verified step by step guidance

Identify the function as a quotient of two functions: $ y = \frac{u}{v} $ where $ u = 2 - 3t $ and $ v = 4t^2 + 7 $.

Recall the Quotient Rule for derivatives, which states: $ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} $.

Calculate $ \frac{du}{dt} $ by differentiating $ u = 2 - 3t $ with respect to $ t $, which gives $ \frac{du}{dt} = -3 $.

Calculate $ \frac{dv}{dt} $ by differentiating $ v = 4t^2 + 7 $ with respect to $ t $, which gives $ \frac{dv}{dt} = 8t $.

Substitute $ u, v, \frac{du}{dt}, \frac{dv}{dt} $ into the Quotient Rule formula: $ \frac{d}{dt}\left(\frac{2-3t}{4t^2+7}\right) = \frac{(4t^2+7)(-3) - (2-3t)(8t)}{(4t^2+7)^2} $.