If 0=4+5x^(3)-y^(2) then find (dy)/(dx) at the point (1,-3). Answer: (dy)/(dx)|_((1,-3))=

Implicit Differentiation: First, we need to implicitly differentiate the given equation with respect to x to find dy/dx. Given equation: 0 = 4 + 5x^3 - y^2 Differentiate both sides with respect to x: d/dx(0) = d/dx(4) + d/dx(5x^3) - d/dx(y^2) Since the derivative of a constant is 0, we have: 0 = 0 + 15x^2 - 2y(dy/dx)Solving for dy/dx: Now, we solve for dy/dx: 0 = 15x^2 - 2y(dy/dx) To isolate dy/dx, we add 2y(dy/dx) to both sides and then divide by 2y: 2y(dy/dx) = 15x^2 (dy/dx) = (15x^2) / (2y)Substitute Given Point: Next, we substitute the given point (1, -3) into the equation to find the value of dy/dx at that point: (dy/dx)|_((1,-3)) = (15(1)^2) / (2(-3)) (dy/dx)|_((1,-3)) = 15 / -6 (dy/dx)|_((1,-3)) = -5/2

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If
0=4+5x^(3)-y^(2) then find
(dy)/(dx) at the point
(1,-3).
Answer:
(dy)/(dx)|_((1,-3))=

If $0=4+5 x^{3}-y^{2}$ then find $\frac{d y}{d x}$ at the point $(1,-3)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(1,-3)}=$

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Math Problems

Grade 8

Evaluate a nonlinear function

Full solution

Q. If

0=4+5 x^{3}-y^{2}

then find

\frac{d y}{d x}

at the point

(1,-3)

\newline

Answer:

\left.\frac{d y}{d x}\right|_{(1,-3)}=

Implicit Differentiation: First, we need to implicitly differentiate the given equation with respect to $x$ to find $\frac{dy}{dx}$ . $\newline$ Given equation: $0 = 4 + 5x^3 - y^2$ $\newline$ Differentiate both sides with respect to $x$ : $\newline$ $\frac{d}{dx}(0) = \frac{d}{dx}(4) + \frac{d}{dx}(5x^3) - \frac{d}{dx}(y^2)$ $\newline$ Since the derivative of a constant is $0$ , we have: $\newline$ $0 = 0 + 15x^2 - 2y\frac{dy}{dx}$
Solving for $\frac{dy}{dx}$ : Now, we solve for $\frac{dy}{dx}$ :
$0 = 15x^2 - 2y\frac{dy}{dx}$
To isolate $\frac{dy}{dx}$ , we add $2y\frac{dy}{dx}$ to both sides and then divide by $2y$ :
$2y\frac{dy}{dx} = 15x^2$
$\frac{dy}{dx} = \frac{15x^2}{2y}$
Substitute Given Point: Next, we substitute the given point $(1, -3)$ into the equation to find the value of $\frac{dy}{dx}$ at that point: $\newline$ $\left(\frac{dy}{dx}\right)|_{(1,-3)} = \frac{15(1)^2}{2(-3)}$ $\newline$ $\left(\frac{dy}{dx}\right)|_{(1,-3)} = \frac{15}{-6}$ $\newline$ $\left(\frac{dy}{dx}\right)|_{(1,-3)} = -\frac{5}{2}$