13-26 Implicit differentiation Carry out the following steps.

Question

a. Use implicit differentiation to find dy/dx.

tan xy = x+y; (0,0)

Accepted Answer

Welcome back, everyone. Given the equation tangent of 5 XY equals 3x minus 7Y, use implicit differentiation to find the first derivative. So let's write down the relationship we're given tangent of 5 XY. is equal to 3 x minus 7 Y. Now for this problem we're going to apply implicit differentiation, and this means that we're going to set the derivative of the left hand side equal to the derivative of the right hand side. So first of all, let's differentiate the tangent function, and we have to recall that the derivative of tangent is c 2d. So we're going to get squared of the original angle 5 x y. But because it's not X, it is a function based on the chain rule, we have to multiply that by the derivative of the inner function, which is 5 xy, and we have to take the derivative of that. Now on the right hand side we have the derivative of 3 X, which gives us 3 based on the power rule. And then we are subtracting 7 multiplied by the derivative of Y, so we get 7 Y. Now, the only term that we are interested in is 5 XY derivative for now, right, because this involves a product and there's nothing else that we can simplify, right? So let's evaluate the derivative of 5 XY and specifically we can just evaluate the derivative of XY because we can factor out the constant. We can say that 5 s can squared of 5 XY multiplied by X Y derivative is equal to 3 minus 7 Y derivative. And now based on the product rule, we have to take x derivative multiplied by Y and add X Y derivative, which gives us Y. the derivative of X is 1 plus xy. So now let's substitute this expression into our equation. We're going to get 5 2nd squared of 5 X Y multiplied by. Y + XY equals 3 minus 7 Y. So that's it. And now we want to solve for Y because this is the Y divided by the X, right? And for that purpose, first of all, let's distribute the terms. We're going to get 5 Y. 2 squared 5 x y plus 5 xy multiplied by squad of 5 X Y is equal to the right hand side 3 minus 7 Y. And now let's group the terms involving Y. We can do that on the left hand side, so we're going to get 5 XY. C squad of 5 xy we're going to add 7 Y to both hand sides, so this gives us plus 7 Y. That'd be more accurate to say right, we have 7 Y. And then on the right hand side, we end up with 3, and let's subtract 5 Y square of 5 XY, which gives us 3 minus 5 Y squad of 5 XY. Now let's factor out Y. Which gives us 5 x 2 squad of 5 Xyres. + 7 equals. We minus 5 Y C squad of 5 XY and now if we want to solve for Y, which is essentially D Y divided by D X, we have to divide. Both sides by the factor adjacent toy. So we take the right hand side 3 minus 5 Y 2 squared off 5 x y divided by. 5 X. Second squared of 5 X Y plus 7. Which corresponds to our final answer. Thank you for watching.

13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
tan xy = x+y; (0,0)

Key Concepts

Implicit Differentiation

Chain Rule

Evaluating at a Point

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