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If
-x^(2)=-5y^(2)+y^(3) then find
(dy)/(dx) at the point
(2,1).
Answer:
(dy)/(dx)|_((2,1))=

If $-x^{2}=-5 y^{2}+y^{3}$ then find $\frac{d y}{d x}$ at the point $(2,1)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(2,1)}=$

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Evaluate a nonlinear function

Full solution

Q. If

-x^{2}=-5 y^{2}+y^{3}

then find

\frac{d y}{d x}

at the point

(2,1)

.

\newline

Answer:

\left.\frac{d y}{d x}\right|_{(2,1)}=

Apply Implicit Differentiation: First, we need to differentiate both sides of the equation with respect to $x$ , using implicit differentiation, because $y$ is a function of $x$ . Differentiate $-x^2$ to get $-2x$ . Differentiate $-5y^2$ with respect to $x$ to get $-10y\frac{dy}{dx}$ . Differentiate $y^3$ with respect to $x$ to get $y$ $0$ . The differentiated equation is: $y$ $1$ .
Factor Out $(\frac{dy}{dx}):</b> Now, we need to solve for \$(\frac{dy}{dx})$ by factoring it out from the terms on the right side of the equation. $\newline$ Group the terms with $(\frac{dy}{dx})$ together: $-2x = (\frac{dy}{dx})(-10y + 3y^2)$ .
Isolate $(\frac{dy}{dx}):</b> Next, isolate \$(\frac{dy}{dx})$ by dividing both sides of the equation by $(-10y + 3y^2)$ . $\newline$ $(\frac{dy}{dx}) = \frac{-2x}{(-10y + 3y^2)}$ .
Substitute Point: Now we need to substitute the point $(2,1)$ into the equation to find the value of $(\frac{dy}{dx})$ at that point. $\newline$ Substitute $x = 2$ and $y = 1$ into the equation: $(\frac{dy}{dx}) = \frac{-2(2)}{(-10(1) + 3(1)^2)}$ .
Calculate (\frac{dy}{dx}): Perform the calculations to find the value of \$(\frac{dy}{dx}).\(\newline\$(\frac{dy}{dx}) = \frac{-4}{(-10 + 3)} = \frac{-4}{(-7)} = \frac{4}{7}.\)

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