27–40. Implicit differentiation Use implicit differentiat...

Question

27–40. Implicit differentiation Use implicit differentiation to find dy/dx.

√x⁴+y² = 5x+2y³

Accepted Answer

Hello. In this video, we are going to be finding the derivative using implicit differentiation. So, we are asked to find the derivative DYDX to the given equation. That equation is the square root of X of 8 plus Y of 5, equal to 9 X plus 7 Y 5th power. So, how can we find the derivative DYDX? Well, through implicit differentiation, we want to go ahead and take the derivative of the dependent variable X. So in order to find DYDX, what we want to do is we want to take the derivative of the entire equation with respect to X. So, let's go ahead and start this process. First, before we take the derivative, let's go ahead and rewrite the square root in its exponential form. That is going to be equivalent to X to the power of 8 plus Y to the power of 5, all raised to the half power. Now, in order to take the derivative of the left hand side, we are going to need to use the chain rule. So, by the chain rule, the outermost derivative of the left hand side is going to be 1/2 multiplied by X to the power of 8 plus Y to the power of 5, raised to the power of -12. Then we have to multiply this derivative by the derivative of the innermost function. Now, we are going to take the derivative of the innermost function with respect to X. X to the power of 8's derivative is going to be 8 multiplied by X rated to the 7th power by the power rule. But what about the derivative of Y to the power of 5? Well, we will still take the derivative by using the power rule, that is going to give us 5 Y to the 4th power. But recall that we are taking a derivative of a function of Y with respect to X. So, by implicit differentiation, that is going to generate a DYDX term. And this is going to be the derivative of the left hand side of the equation. Now we will take the derivative of the right-hand side of the equation with respect to X. The derivative of 9 X is going to be 9 by the power rule. And again, we will take the derivative of 7 to 5th with respect to X. That is going to give us 7 multiplied by 5 to the power of 4, and because we took the derivative of Y with respect to X, we will generate a DYDX term. Now let's go ahead and simplify. Since we want to work with positive exponents, we will move X to the power of 8 plus Y to the power of 5, raise the negative half power to the denominator of 1/2 to make the exponent positive. Then we will multiply everything by 8X7 plus YY4 DYD X. That will allow us to simplify the left hand side of the equation. To be 8 X to the 7th power, plus 5 to the 4th power, DYDX all divided. By 2 multiplied by the square root of X to the power of 8 plus Y to the power of 5. On the right hand side, we will multiply 7 with 5 to the 4th power. That is going to simplify the right-hand side of the equation to be 9 plus 35 to 4th DYDX. So, this is the derivative of the entire equation with respect to X, but our goal is to solve for the derivative DYDX. So what that means is that we want to try to get all of our terms that include a DYDX term to one side of the equation in order to solve. So how do we start this process? Well, first, let's go ahead and eliminate the denominator from the left-hand side of the equation. We will multiply both sides of the equation by 2, multiply it by square root of X to the power of 8, plus Y to the power of 5. That will allow us to rewrite the the derivative as 8 X to the 7th power, plus 5 to 4th power, DYDX equal to 2 square root of X 8 plus Y 5th multiplied by the quantity, 9 plus 35 to the 4th, DYDX. Now, what we want to do is we want to distribute the square root term into the quantity 9 + 354 DYDX. That will give us 8 X to the 7th, plus 5 to the 4th, DYDX equal to 18 square root of X 8 plus the 5th. Plus 35 multiplied by 2, which is going to give us 70. Square root X to the 8th, Y to the 5th. Multiplied by the 4th, DYD X. Now that we have everything in the form of a sum, we are going to move all the DYDX terms to one side of the equation, and all the constants to the other side of the equation. So what we can go ahead and do is let's go ahead and subtract 5Y to the 4th DYDX to one side of the equation. And subtract 18 square root X to the 8th to the 5th to the other side. That will leave us with 8 X 7 minus 18 multiplied by X 8 plus Y 5th, and that will equal to 70 square root. X 8 plus the 5th. Multiplied by the 4th, DYDX minus 5 Y 4th, DYDX. Notice that both of the terms on the right-hand side of the equation have a DYDX term. So what we can go ahead and do is factor out the DYDX from the right-hand side of the equation. That will allow us to simplify the right-hand side to be 704 square root of X 8 plus Y 5th, minus 5 to 4th power all multiplied by DYDX, and this is still going to equal to 8 X 7th minus 18 multiplied by X 8 plus Y 5th. Now, in order to solve for DYDX, we are going to divide both sides of the equation. By 70 to 4th square root of X 8 plus the 5th, minus 5 to 4th. That will leave us with D Y D X equal to 70 oops sorry. That will be numerator 8X to the 7th, minus 18 multiplied by the square root of X 8 plus Y 5th, all divided. By 70, to the 4th square root. X 8 plus the 5th, minus 5 to 4th power. Now there is one more thing that we can do to simplify the derivative. In the numerator, we can go ahead and factor out. A 2, and we can factor out from the denominator, a 5 to the 4th power. That is going to simplify the derivative to be 2 multiplied by the quantity, 4 X. To the 7th power. Minus 9 multiplied by the square root of X to the 8th plus Y 5th divided by 5 to 4th power multiplied by 14. Square root of x 8 plus y 5th. -1. And this is going to be the simplest form of the DYDX derivative. So I hope this video helps you in understanding how to implicitly differentiate an equation, and I will go ahead and see you all in the next video.

27–40. Implicit differentiation Use implicit differentiation to find dy/dx.√x⁴+y² = 5x+2y³

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27–40. Implicit differentiation Use implicit differentiation to find dy/dx.
√x⁴+y² = 5x+2y³