Question

find \\( \frac { d y } { d x } \\) by implicit differentiation.
\\( y = 4 x ^ { 2 } + 9 y x \\)
\\( \frac { d y } { d x } = \square \\)

Explanation:

Step1: Differentiate both sides with respect to \(x\)

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

Step2: Solve for \(\frac{dy}{dx}\)

Rearrange the terms to isolate \(\frac{dy}{dx}\) on one side.
\(\frac{dy}{dx}-9x\frac{dy}{dx}=8x + 9y\).
Factor out \(\frac{dy}{dx}\) on the left - hand side: \(\frac{dy}{dx}(1 - 9x)=8x + 9y\).
Then \(\frac{dy}{dx}=\frac{8x + 9y}{1 - 9x}\).

Answer:

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