If 5y=-xy^(2)-x 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 have the equation 5y = -xy^2 - x. To find (dy)/(dx), we need to differentiate both sides of the equation with respect to x, using the product rule for the term -xy^2 and remembering that y is a function of x. Differentiating the left side: d/dx (5y) = 5 * (dy)/(dx) Differentiating the right side: d/dx (-xy^2 - x) = d/dx (-xy^2) + d/dx (-x) For the first term, we use the product rule: d/dx (uv) = u'v + uv' Let u = x and v = y^2, then u' = 1 and v' = 2y(dy)/(dx) So, d/dx (-xy^2) = -1 * y^2 - x * 2y * (dy)/(dx) And d/dx (-x) = -1 Combining the derivatives, we get: - y^2 - 2xy * (dy)/(dx) - 1 Now we equate the derivatives of both sides: 5 * (dy)/(dx) = - y^2 - 2xy * (dy)/(dx) - 1Derivative calculation: Solve for (dy)/(dx). We now have an equation with (dy)/(dx) on both sides: 5 * (dy)/(dx) = - y^2 - 2xy * (dy)/(dx) - 1 To solve for (dy)/(dx), we need to get all the terms involving (dy)/(dx) on one side of the equation: 5 * (dy)/(dx) + 2xy * (dy)/(dx) = - y^2 - 1 Factor out (dy)/(dx): (dy)/(dx) * (5 + 2xy) = - y^2 - 1 Now, divide both sides by (5 + 2xy) to isolate (dy)/(dx): (dy)/(dx) = (- y^2 - 1) / (5 + 2xy)

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

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

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Q. If

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

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 have the equation $5y = -xy^2 - x$ . To find $\frac{dy}{dx}$ , we need to differentiate both sides of the equation with respect to $x$ , using the product rule for the term $-xy^2$ and remembering that $y$ is a function of $x$ . Differentiating the left side: $\frac{d}{dx} (5y) = 5 \cdot \frac{dy}{dx}$ Differentiating the right side: $\frac{d}{dx} (-xy^2 - x) = \frac{d}{dx} (-xy^2) + \frac{d}{dx} (-x)$ For the first term, we use the product rule: $x$ $0$ Let $x$ $1$ and $x$ $2$ , then $x$ $3$ and $x$ $4$ So, $x$ $5$ And $x$ $6$ Combining the derivatives, we get: $x$ $7$ Now we equate the derivatives of both sides: $x$ $8$
Derivative calculation: Solve for $(dy)/(dx)$ . $\newline$ We now have an equation with $(dy)/(dx)$ on both sides: $\newline$ $5 \cdot (dy)/(dx) = - y^2 - 2xy \cdot (dy)/(dx) - 1$ $\newline$ To solve for $(dy)/(dx)$ , we need to get all the terms involving $(dy)/(dx)$ on one side of the equation: $\newline$ $5 \cdot (dy)/(dx) + 2xy \cdot (dy)/(dx) = - y^2 - 1$ $\newline$ Factor out $(dy)/(dx)$ : $\newline$ $(dy)/(dx) \cdot (5 + 2xy) = - y^2 - 1$ $\newline$ Now, divide both sides by $(5 + 2xy)$ to isolate $(dy)/(dx)$ : $\newline$ $(dy)/(dx)$ $0$