Find the value of dy/dx at the point (5, −2) for the equation 3×2 - 5y3 + xy - 1 = 0.

Question

ScanSolve AI Tutor · Accepted Answer

Here's the solution for the first problem.

Differentiate the given equation implicitly with respect to $x$ .
The equation is $3x^2 - 5y^3 + xy - 1 = 0$ .
Differentiating each term:
$\frac{d}{dx}(3x^2) - \frac{d}{dx}(5y^3) + \frac{d}{dx}(xy) - \frac{d}{dx}(1) = \frac{d}{dx}(0)$ $6x - 15y^2 \frac{dy}{dx} + \left(1 \cdot y + x \cdot \frac{dy}{dx}\right) - 0 = 0$ $6x - 15y^2 \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$
Group terms with $\frac{dy}{dx}$ and solve for $\frac{dy}{dx}$ .
$x \frac{dy}{dx} - 15y^2 \frac{dy}{dx} = -6x - y$ $\frac{dy}{dx}(x - 15y^2) = -6x - y$ $\frac{dy}{dx} = \frac{-6x - y}{x - 15y^2}$
Substitute the given point $(5, -2)$ into the expression for $\frac{dy}{dx}$ .
Here, $x = 5$ and $y = -2$ .
$\frac{dy}{dx} = \frac{-6(5) - (-2)}{5 - 15(-2)^2}$ $\frac{dy}{dx} = \frac{-30 + 2}{5 - 15(4)}$ $\frac{dy}{dx} = \frac{-28}{5 - 60}$ $\frac{dy}{dx} = \frac{-28}{-55}$ $\frac{dy}{dx} = \frac{28}{55}$
The value of $\frac{dy}{dx}$ at the point $(5, -2)$ is $\boxed{\frac{28}{55}}$ .

Find the value of dy/dx at the point (5, -2) for the equation 3x2 − 5y3 + xy − 1 = 0.