Derivatives in Differential Form

Question

In Exercises 17–28, find dy.

y = x√(1 − x²)

Accepted Answer

Welcome back, everyone. In this problem, we want to find D Y for the equation Y E equals 5X multiplied by the square root of 121 minus X2. Now, how are we going to find the Y? What do we know here if we want to differentiate Y? Well, notice here that we have a product, OK? And to differentiate products, we can use the product rule. Recall that the product rule basically tells us that if we have a function Y that's equal to UV where U and V are functions of X, then the derivative of Y with respect to X is going to be equal to. It's gonna be equal to uh you multiplied by DVDX. Plus V multiplied by DUD X. In other words, we differentiate U and V and then multiply them by the opposite term. But now, remember, in this case, since we're solving for D Y and each term in this equation is being divided by D X, we can divide through by DX to get D Y to be equal to UDV plus VDU. So if we can differentiate U and V and use that to solve for a DV and DU, then we should be able to solve for DY. So let's go ahead and try to do that. First, let's let U be equal to 5 X. And let's let V be equal to the square root of 121. Minus X2. That means DUDX is going to be equal to 5, which means DU is gonna be equal to 5DX. On the other hand, DVDX we can differentiate it by the chain rule, and it is gonna be equal. So the derivative of our uh alter function, so that's 1 divided by 2 multiplied by root 121 minus X2, multiplied by the derivative of our inner function, it's -2 X, OK, which is gonna be equal to negative, well, 22, so it's gonna be negative X divided by the square root of 121 minus X2. If that's the case, then that means DV is gonna be equal to negative X. Multiplied by the square root of 121 minus X squared multiplied by the X. Now that we have DU and the DV, we can substitute them into the equation that we found earlier, OK? Which now tells us then that Dy is gonna be equal to, like we said, U 5 X multiplied by DV. Negative X divided by the square root of 121 minus X 2 DX, OK. Plus V, uh, which is, well, we said V is the square root of 121 minus X squared multiplied by DU which is 5DX. Now, we can do some simplification here, OK? Because now, in our first term, we could rewrite this as negative 5X2d divided by the square root of 121 minus X2 DX. And for a second term, it's gonna be 5 multiplied by the skirt of 121 minus X squared DX. And notice that since both terms have DX we can factor out the X, OK, which means DY is gonna be negative 5X2 divided by the square root of 121 minus X2 plus 5 multiplied by the square of 121 minus X2 DX. Now we can simplify even further by trying to combine our terms, OK? So let's try to get a common denominator here. No, in this case, we could rewrite this, uh, sharing the denominator of the square root of 121 minus X squared. Then this is gonna be 5 multiplied by the square root of 121 minus X squared all squared, which is just gonna give us 121 minus X2. OK. Minus 5 X squad, OK, all divided by the square root of 121 minus X2 X. I know if we expand in our first term, OK, that's gonna be 5, OK. Multiplied by 121 minus X2 minus X2. OK. Keeping our CM denominator. And now if we take it a step further, minus X2 minus X is -2 X. This is gonna be 5 multiplied by 121 minus 2 X2. OK, all divided by the square root of 121 minus X 2 DX. This is the differential D Y. Thanks a lot for watching everyone. I hope this video helped.

Derivatives in Differential Form

In Exercises 17–28, find dy.

y = x√(1 − x²)

Key Concepts

Derivative

Product Rule

Chain Rule

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