Find the derivative of the function.g(y)=5y2+2y−1y2−2g(y)=\frac{5y^2+2y-1}{y^2-2}g(y)=y2−25y2+2y−1

− 2 y 2 − 18 y − 4 ( y 2 − 2 ) 2 \frac{-2y^2-18y-4}{(y^2-2)^2} ( y 2 − 2 ) 2 − 2 y 2 − 18 y − 4

Find the derivative of the function. g ( y ) = 5 y 2 + 2 y − 1 y 2 − 2 g(y)=\frac{5y^2+2y-1}{y^2-2} g ( y ) = y 2 − 2 5 y 2 + 2 y − 1

this problem, we're asked to find the derivative of the function g of y equals 5 y squared plus 2 y minus 1 divided by y squared minus 2. So I have one function being divided by another function, and we have to use, you guessed it, the quotient rule to find this derivative. So applying our quotient rule to find our derivative here, g prime of y, remember that when we use the quotient rule, we have 2 functions. We have our top function that we think of as our high function and we have our bottom function that is our low function so that we can remember our rhyme here, low d high minus high d low over the square of what's below to find this derivative here. So starting with that first term, low d high, I'm gonna take that bottom function, my low function, y squared minus 2 times the derivative of that top function. Now the derivative of 5 y squared plus 2 y minus 1, that is going to give me 10y+2 using my power rule here. Remember that the derivative of any constant is just 0. That's why I only have 2 terms here and not 3. Now I'm subtracting here high d low, so taking that high function leaving it as is, 5y squared plus 2 y minus 1, and then multiplying that by the derivative of that bottom function, y squared minus 2. The derivative of y squared minus 2 is just 2 y, and that is our full numerator. Now we are dividing all of this by the square of what's below. So that's a y squared minus 2. And remember to square that. Now from here, we have applied our quotient rule, and you know the drill. It is all algebra from here. So let's go ahead and simplify this using our algebra skills. Now looking at this first term in our numerator we have these two polynomials multiplying each other so I'm going to go ahead and foil this out. So I'm going to take those first two terms y squared and ten y. Multiplying those gives me 10 y cubed. Then I'm going to take those outside terms. That's y squared and 2. That's going to give me plus 2 y squared. Then my inside terms, negative two times 10 y. That's going to give me negative 20 y. And then my last terms, negative 2 times positive 2, that gives me negative 4. Now we have quite a bit of algebra going on here because that is just our first term in our numerator. Remember, we have our second term here that is all being subtracted. I'm going to keep it in parentheses for now so that I remember to apply this negative to all of those terms. So distributing this 2y into my parentheses here that's going to give me 10y cubed plus 4y minus or plus 4y squared, excuse me, minus 2y. Now from here, I am dividing by the same thing. That denominator is staying the same, y squared minus 2 squared. And of course, we want to distribute that negative in here. I'm not going to rewrite this. I'm just going to apply that negative that I already know. So I have negative 10 y cubed and then this becomes a negative as well. And then that last negative becomes a positive because we know 2 negatives makes a positive. So now from here I have successfully applied all of that algebra, but we do have some like terms. So I have 10 y cubed and I have a negative 10 y cubed, which are going to cancel each other out. Then I have 2 y squared minus 4 y squared. That is going to give me negative 2 y squared. Then I have negative 20 y and positive 2 y. That's going to give me negative 18 y. And then I have this constant minus 4. It's really important to stay organized and make sure that you're accounting for every single term here. Then my denominator stays as is. It's that y squared minus 2 squared. And this is our final derivative here at g prime of y simplified as much as we can. Feel free to let me know if you have any questions here. And let's get some more practice coming up.

Find the derivative of the function. g ( y ) = 5 y 2 + 2 y − 1 y 2 − 2 g(y)=\frac{5y^2+2y-1}{y^2-2} g ( y ) = y 2 − 2 5 y 2 + 2 y − 1

− 2 y 2 − 18 y − 4 ( y 2 − 2 ) 2 \frac{-2y^2-18y-4}{(y^2-2)^2} ( y 2 − 2 ) 2 − 2 y 2 − 18 y − 4

2 y 2 + 18 y + 4 ( y 2 − 2 ) 2 \frac{2y^2+18y+4}{(y^2-2)^2} ( y 2 − 2 ) 2 2 y 2 + 18 y + 4

2 y 2 + 18 y + 4 ( 5 y 2 + 2 y − 1 ) 2 \frac{2y^2+18y+4}{(5y^2+2y-1)^2} ( 5 y 2 + 2 y − 1 ) 2 2 y 2 + 18 y + 4

− 2 y 2 − 18 y ( y 2 − 2 ) 2 \frac{-2y^2-18y}{(y^2-2)^2} ( y 2 − 2 ) 2 − 2 y 2 − 18 y

3. Techniques of Differentiation

Product and Quotient Rules

Business Calculus

3. Techniques of Differentiation

Product and Quotient Rules

Struggling with Business Calculus?

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Multiple Choice

Find the derivative of the function.
$g(y)=\frac{5y^2+2y-1}{y^2-2}$

$\frac{2y^2+18y+4}{(y^2-2)^2}$

$\frac{-2y^2-18y-4}{(y^2-2)^2}$

$\frac{2y^2+18y+4}{(5y^2+2y-1)^2}$

$\frac{-2y^2-18y}{(y^2-2)^2}$

Verified step by step guidance

Step 1: Recognize that the given function g(y) is a rational function, meaning it is a fraction where both the numerator and denominator are polynomials. To find the derivative of g(y), we will use the Quotient Rule.

Step 2: Recall the Quotient Rule formula: If g(y) = f(y)/h(y), then g'(y) = (f'(y)h(y) - f(y)h'(y)) / [h(y)]^2. Here, f(y) = 5y^2 + 2y - 1 (the numerator) and h(y) = y^2 - 2 (the denominator).

Step 3: Compute the derivative of the numerator, f'(y). For f(y) = 5y^2 + 2y - 1, use the power rule: f'(y) = d/dy[5y^2] + d/dy[2y] - d/dy[1]. This simplifies to f'(y) = 10y + 2.

Step 4: Compute the derivative of the denominator, h'(y). For h(y) = y^2 - 2, use the power rule: h'(y) = d/dy[y^2] - d/dy[2]. This simplifies to h'(y) = 2y.

Step 5: Substitute f(y), f'(y), h(y), and h'(y) into the Quotient Rule formula. Simplify the numerator of the derivative expression, which is (f'(y)h(y) - f(y)h'(y)), and divide by [h(y)]^2. The final expression will be the derivative of g(y).

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