f(x,y)=x^3+y^2x-3 find dy/dx

Question

nick1337 · Accepted Answer

Differentiate both sides of the equation.

$\frac{d}{d x} (f (x, y)) = \frac{d}{d x} (x^{3} + y^{2} x - 3)$

Since $f$ is constant with respect to $x$ , the derivative of $f (x, y)$ with respect to $x$ is $f \frac{d}{d x} [(x, y)]$ .

$f \frac{d}{d x} [(x, y)]$

Differentiate the right side of the equation.

$3 x^{2} + \frac{d}{d x} [y^{2} x] + \frac{d}{d x} [- 3]$

Evaluate $\frac{d}{d x} [y^{2} x]$ .

$3 x^{2} + y^{2} + 2 x y y' + \frac{d}{d x} [- 3]$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$3 x^{2} + y^{2} + 2 x y y' + 0$

Add $3 x^{2} + y^{2} + 2 x y y'$ and $0$ .

$3 x^{2} + y^{2} + 2 x y y'$

Reform the equation by setting the left side equal to the right side.

$f \frac{d}{d x} ((x, y)) = 3 x^{2} + y^{2} + 2 x y y'$

Solve for $y'$ .

Rewrite the equation as $3 x^{2} + y^{2} + 2 x y y' = \frac{f d}{d x ((x, y))}$ .

$3 x^{2} + y^{2} + 2 x y y' = \frac{f d}{d x ((x, y))}$

Cancel the common factor of $d$ .

$3 x^{2} + y^{2} + 2 x y y' = \frac{f}{x ((x, y))}$

Move all terms not containing $y'$ to the right side of the equation.

$2 x y y' = \frac{f}{x ((x, y))} - 3 x^{2} - y^{2}$

Divide each term in $2 x y y' = \frac{f}{x ((x, y))} - 3 x^{2} - y^{2}$ by $2 x y$ and simplify.

$y' = \frac{f}{2 x^{2} y (x, y)} - \frac{3 x}{2 y} - \frac{y}{2 x}$

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{f}{2 x^{2} y (x, y)} - \frac{3 x}{2 y} - \frac{y}{2 x}$

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