If -5+x^(2)-y^(2)+y=0 then find (dy)/(dx) in terms of x and y. Answer: (dy)/(dx)=

Differentiate with respect to x: Differentiate both sides of the equation with respect to x. We will use implicit differentiation because y is a function of x. Differentiate each term separately: The derivative of -5 with respect to x is 0. The derivative of x^2 with respect to x is 2x. The derivative of -y^2 with respect to x is -2y(dy/dx) because we apply the chain rule. The derivative of y with respect to x is (dy/dx). So, we have: 0 + 2x - 2y(dy/dx) + (dy/dx) = 0Apply chain rule: Solve for (dy/dx). We will collect the terms with (dy/dx) on one side and the rest on the other side: -2y(dy/dx) + (dy/dx) = -2x Now, factor out (dy/dx): (dy/dx)(-2y + 1) = -2x Now, divide both sides by (-2y + 1) to solve for (dy/dx): (dy/dx) = -2x / (-2y + 1)

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If
-5+x^(2)-y^(2)+y=0 then find
(dy)/(dx) in terms of
x and
y.
Answer:
(dy)/(dx)=

If $-5+x^{2}-y^{2}+y=0$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$

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Math Problems

Algebra 1

Transformations of quadratic functions

Full solution

Q. If

-5+x^{2}-y^{2}+y=0

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate with respect to $x$ : Differentiate both sides of the equation with respect to $x$ . We will use implicit differentiation because $y$ is a function of $x$ . Differentiate each term separately: The derivative of $-5$ with respect to $x$ is $0$ . The derivative of $x^2$ with respect to $x$ is $2x$ . The derivative of $x$ $0$ with respect to $x$ is $x$ $2$ because we apply the chain rule. The derivative of $y$ with respect to $x$ is $x$ $5$ . So, we have: $x$ $6$
Apply chain rule: Solve for $(\frac{dy}{dx})$ . We will collect the terms with $(\frac{dy}{dx})$ on one side and the rest on the other side: $-2y(\frac{dy}{dx}) + (\frac{dy}{dx}) = -2x$ Now, factor out $(\frac{dy}{dx})$ : $(\frac{dy}{dx})(-2y + 1) = -2x$ Now, divide both sides by $(-2y + 1)$ to solve for $(\frac{dy}{dx})$ : $(\frac{dy}{dx}) = \frac{-2x}{-2y + 1}$