Question

find $\frac{dy}{dx}$ by implicit differentiation.
$y = 4x^{2}+9yx$
$\frac{dy}{dx}=\frac{8x - 9y}{1 - 9x}$
thats incorrect.
to use implicit differentiation, first differentiate both sides of the equation with
respect to $x$, treating $y$ as a differentiable function of $x$. then, collect the terms with
$\frac{dy}{dx}$ on one side of the equation. finally, solve for $\frac{dy}{dx}$.

Explanation:

Step1: Differentiate both sides

Differentiate \(y = 4x^{2}+9yx\) with respect to \(x\).
Using the sum rule \((u + v)^\prime=u^\prime + v^\prime\), where \(u = 4x^{2}\) and \(v=9yx\).
The derivative of \(4x^{2}\) with respect to \(x\) is \(8x\) (using the power rule \((x^{n})^\prime=nx^{n - 1}\)).
For the term \(9yx\), use the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = 9y\) and \(v=x\). So \((9yx)^\prime=9\frac{dy}{dx}\cdot x+9y\cdot1 = 9x\frac{dy}{dx}+9y\).
The derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx}\). So we have \(\frac{dy}{dx}=8x + 9x\frac{dy}{dx}+9y\).

Step2: Collect \(\frac{dy}{dx}\) terms

Subtract \(9x\frac{dy}{dx}\) from both sides:
\(\frac{dy}{dx}-9x\frac{dy}{dx}=8x + 9y\).
Factor out \(\frac{dy}{dx}\) on the left - hand side: \(\frac{dy}{dx}(1 - 9x)=8x + 9y\).

Step3: Solve for \(\frac{dy}{dx}\)

Divide both sides by \((1 - 9x)\) (assuming \(x
eq\frac{1}{9}\)): \(\frac{dy}{dx}=\frac{8x + 9y}{1 - 9x}\).

Answer:

\(\frac{dy}{dx}=\frac{8x + 9y}{1 - 9x}\)