Solve the equation explicitly for y and differentiate to get y’ in terms of x. 1x+1y=1

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Elberte · Accepted Answer

1x+1y=1  1y=1−1x  1y=x−1x  Take reciporal of both sides  y=xx−1  Differentiate both sides with respect to x  dydx=ddx[xx−1]  dydx=(x−1)⋅dxdx−x⋅d(x−1)dx(x−1)2⏟QUOTIENT RULE  dydx=(x−1)⋅1−x⋅1(x−1)2  dydx=x−1−x(x−1)2  dydx=−1(x−1)2  Result: dydx=−1(x−1)2

user_27qwe · Accepted Answer

To solve the equation 1x+1y=1 explicitly for y and differentiate to get y′ in terms of x, we can follow these steps:Step 1: Rearrange the equation to isolate y:1y=1−1xStep 2: Take the reciprocal of both sides:y=11−1xStep 3: Simplify the expression on the right-hand side by finding a common denominator:y=1x−1xStep 4: Invert the fraction and multiply:y=xx−1Now, to differentiate y with respect to x, we can use the quotient rule. Let&#039;s denote the derivative of y with y′:y′=ddx(xx−1)Using the quotient rule, we have:y′=(x−1)ddx(x)−xddx(x−1)(x−1)2Simplifying this expression gives us:y′=(x−1)(1)−x(1)(x−1)2Further simplification yields:y′=x−1−x(x−1)2y′=−1(x−1)2Hence, the solution to the equation and the derivative y′ are:y=xx−1y′=−1(x−1)2

Vasquez · Accepted Answer

Answer:y′=−1(x−1)2Explanation:Step 1: Begin by subtracting 1x from both sides of the equation:1y=1−1xStep 2: Take the reciprocal of both sides to isolate y:y=11−1xStep 3: Simplify the expression on the right side by finding a common denominator:y=1xx−1x=1x−1x=xx−1Therefore, the equation explicitly solved for y is:y=xx−1Step 4: To find y′, we need to differentiate y with respect to x. Using the quotient rule, we have:y′=ddx(xx−1)=(x−1)ddx(x)−xddx(x−1)(x−1)2Simplifying further:y′=(x−1)(1)−x(1)(x−1)2=x−1−x(x−1)2=−1(x−1)2Hence, y′ in terms of x is:y′=−1(x−1)2

karton · Accepted Answer

Step 1: Solving for yWe&#039;ll start by rearranging the equation to isolate y on one side:1y=1−1xNext, we&#039;ll take the reciprocal of both sides to solve for y:y=11−1xStep 2: Differentiating to find y′ in terms of xTo differentiate y with respect to x, we&#039;ll use the chain rule. Let&#039;s denote 11−1x as u, and we&#039;ll find dudx:u=11−1xTo differentiate u, we&#039;ll apply the chain rule:dudx=ddx(11−1x)Using the quotient rule, the derivative of u can be found as:dudx=−1(1−1x)2·(ddx(1−1x))The derivative of (1−1x) can be calculated as:ddx(1−1x)=0−(1x2)=−1x2Plugging this value back into the equation, we have:dudx=−1(1−1x)2·(−1x2)Simplifying further, we get:dudx=1x2(1−1x)2Finally, substituting back u as 11−1x, we obtain:dudx=1x2(1−1x)2Therefore, the derivative y′ in terms of x is given by:y′=1x2(1−1x)2To summarize, the solution to the equation 1x+1y=1 is given by y=11−1x, and the derivative of y with respect to x is y′=1x2(1−1x)2.

Solve the equation explicitly for y and differentiate to get y’ in terms of x. P

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