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Solve the given problems by finding the appropriate derivatives.In the design of a rectangular container the area $A$ (in $\mathrm{cm}^{2}$ ) of the base is expressed as $A=6 x^{2}-11 x-10,$ and the height $h$ (in cm) is $h=4 x+3 .$ Use the product rule to find the derivative of the volume $V$ with respect to $x$ for $x=5.00 \mathrm{cm}$.

Calculus 1 / AB

Chapter 23

The Derivative

Section 6

Derivatives of Products and Quotients of Functions

Derivatives

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And it seems like a very weird way of asking this question because you can do it a lot faster. I think you can do it a lot faster by doing something else minus 11 x minus 10. Um Well I guess I guess what we're doing is we're finding the volume and the area. Uh Yeah, yeah, I'm not sure of the problem. And the height is this four x minus three. Four X plus three, sorry. So the volume is multiplying those two pieces together. So yeah, we're going to use the product rule because right here is your product to find D. V D X. Which you take the derivative of the left side, which would be 12 x minus 11. Leave the right side alone. And then plus now take the derivative of the right side, which is for leaving the right side alone. And now what we're asked to do, so I don't know if you want to circle that, but that's the correct derivative, is they're asking you to figure out the answer when ax equals five. And uh some students might just go to a calculator. Um because I don't know how good you'd be at at knowing 12 times five is 60 um minus 11, so I'll be 49 then four times five is 23 plus for Yeah, you might want a calculator to get all these answers, and that's what I'm doing. Uh six times 25 because that's what five squared is 11 times five is 55 minus 10, it is 85. So then from here, I definitely want to calculate because I don't feel like doing 49 times 23 in my head or uh four times 85. But the correct answer should be 14 67 as far as units go. Um It looks like that since centimetres and I don't know, it doesn't give us really an X. What does X. Stand for? I don't know. It's usually a time unit. I don't know at the time as though when I'm rereading the problem. All right, so I'm just going to have 14 67 and down.

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