AB MC 3 Q13 18

Added: 2026-05-09

[00:00:00]number 13 says which of the following expressions can be differentiated using the product rule so a says cosine of the square root of x of again means chain rule b says x squared times the arctangent of x x squared times the arc tangent of x is a product rule c says x to the fourth plus arc sine of x that's a sum that isn't any of our rules and finally d says eight x cubed minus five x plus two to the pi that's a chain rule okay number 14 says which of the following requires the use of implicit differentiation implicit differentiation is necessary when you can't get to y equals only so if you look this would be an easy one to get to y equals you just move everything to the other side b is already y equals that's an easy one d is already y equals that's an easy one c cannot really be solved for y because y is in multiple places that would be really difficult to solve for y that is why we have implicit differentiation number 15 says for which the values would the quotient rule be the best method so the quotient rule means we need to have division in a there isn't division of any kind that's a chain rule so the quotient rule isn't necessary for b it looks like a quotient but when you only have one thing on the bottom when there's only one on the bottom it's easier to split the fraction and cancel things down and then it's no longer a quotient rule remember we don't like the quotient rule if we can avoid it c is an arc sine of something that it's only a chain rule there's no division d has division and can't be split up that's going to be your arc or your quotient rule rather number 16 says if y equals 2 natural log of x then the fourth derivative is equal to well 2 natural log of x the 2 stays in natural log of x becomes 1 over x now we're going to take the derivative for the rest of these using the power rule this is x to the negative 1 so it becomes negative 2 x to the negative 2. x to the negative 2 becomes positive 4 x to the negative third x to the negative third becomes negative 12 x to the negative fourth and that is b number 17 says the figure above shows the graph of the derivative and it asks questions about the second derivative so where is the second derivative greatest is wherever the first derivative's slope is the highest so this one has kind of a flat positive slope this one has a slope of zero this one has a very negative slope the largest numerical value would be this one as it is a steep positive slope and then last for this assignment number 18 says let y be a differentiable function a twice differentiable function such that f of one is three and d y d x equals four square root of y squared plus seven x squared what is the value of the second derivative at 1 so the first thing i did is i plugged in 1 and 3 to this 1 squared times 7 is 7 3 squared is 9 7 and 9 is 16 so that's the square root of 16 or 4 times another 4 means d y d x in this problem is equal to 16. we're going to go ahead and take the second derivative now the second derivative is the 4 out front this is the one-half power so one-half keep everything inside the same y squared plus 7x squared to the negative one-half times the chain rule says the derivative of the inside the derivative of the inside is 2y from y squared and dydx because of y plus 14x from 7x squared now it's just a matter of plugging in a bunch of numbers i simplified a little bit first i brought the negative one half down to the bottom i said four times one half was two this is still on the top and then i just started plugging in the different values so two times two times three times we said this was 16 dydx is 16 plus 14 times 2 or i'm sorry 1 x was 1 over we already know this square root is 16 from over here so we multiply all that together you get 220 over 4 220 over 4 gives me an answer of 55.