(a) Find y^' by implicit differentiation. (b) Solve the...

(a) Find $y^{\prime}$ by implicit differentiation. (b) Solve the equation explicitly for $y$ and differentiate to get $y^{\prime}$ in terms of $x .$ (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for $y$ into your solution for part (a). $$2 x^{2}+x+x y=1$$

(a) Find $y^{\prime}$ by implicit differentiation.
(b) Solve the equation explicitly for $y$ and differentiate to get $y^{\prime}$ in terms of $x .$
(c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for $y$ into your solution for part (a).
$$2 x^{2}+x+x y=1$$

Key Concepts

Implicit Differentiation

This concept involves differentiating an equation in which the variable y is defined as an implicit function of x, without first solving for y. It requires applying the chain rule to terms containing y, treating y as a function of x, and then solving for the derivative y?.

Product Rule

The product rule is a fundamental differentiation technique used when differentiating the product of two functions. In the context of implicit differentiation, it is applied to terms like x·y, differentiating it as the derivative of x times y plus x times the derivative of y.

Explicit Differentiation after Solving for y

This concept refers to first isolating y in terms of x from an equation, so that y is expressed explicitly. Once the explicit form is obtained, standard differentiation techniques can be applied directly to find y?, serving as an alternative method to implicit differentiation.

Verification of Differentiation Methods

Verifying the consistency between derivative results obtained through different methods (implicit and explicit differentiation) is an important process in calculus. This verification helps confirm that both approaches yield equivalent derivatives and enhances understanding of the relationship between implicit and explicit forms of functions.

Transcript

00:01 For a, we want to find d .y over d .x or the derivative of y.

00:06 And to do this, we can take our equation and differentiate both sides of the equation in respect to x.

00:21 Okay, starting on the left side, 2x squared, the derivative of that is going to be 4x plus the derivative of x, which is 1.

00:33 Now we have a product rule.

00:37 The derivative of x is also 1, which is y plus x, x.

00:44 X times the derivative y, which is 1.

00:47 But since we're deriving in respect to x, it has to be multiplied by y prime, is all equal to the derivative of 1, which is 0.

00:58 Okay.

01:00 Now what we want to do is try to isolate our y prime variable.

01:04 So what we're going to do is subtract everything from this side of the equation that is not multiplied by y prime to the upside of 3 by 1 and subtract by y.

01:26 Okay, you should end up with this.

01:28 And now we're going to divide both sides by x to isolate that y.

01:31 Prime variable okay this is your answer for the derivative of y now for b what we want to do is first solve the equation explicitly for y this just means moving around the equation to solve for y just algebraically so we're going to do in our first equation here we're going to subtract both sides by subtract both sides by negative 2x squared and subtract both sides by x2 to isolate that x y okay, you should end up with this.

02:30 And now we're going to divide both sides by to isolate that y.

02:43 Okay.

02:45 Now what we want to do from here is differentiate this equation because we do that.

02:50 We'll just put the y prime variable immediately when we just derive this y.

02:56 I'll do it right here...

Please give Ace some feedback