Question

use implicit differentiation to find $\frac{dy}{dx}$.
$x^{6}-48xy + y^{6}=1$
$\frac{dy}{dx}=\square$

Explanation:

Step1: Differentiate each term

Differentiate \(x^{6}-48xy + y^{6}=1\) term - by - term with respect to \(x\).
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), \(\frac{d}{dx}(x^{6}) = 6x^{5}\).
For the term \(-48xy\), use the product rule \(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\), where \(u=-48x\) and \(v = y\). So \(\frac{d}{dx}(-48xy)=-48y-48x\frac{dy}{dx}\).
Using the chain rule \(\frac{d}{dx}(y^{n})=ny^{n - 1}\frac{dy}{dx}\), \(\frac{d}{dx}(y^{6})=6y^{5}\frac{dy}{dx}\), and \(\frac{d}{dx}(1) = 0\).
The derivative of the left - hand side is \(6x^{5}-48y-48x\frac{dy}{dx}+6y^{5}\frac{dy}{dx}\), and the derivative of the right - hand side is \(0\). So we have the equation \(6x^{5}-48y-48x\frac{dy}{dx}+6y^{5}\frac{dy}{dx}=0\).

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

Group the terms with \(\frac{dy}{dx}\) together:
\(6y^{5}\frac{dy}{dx}-48x\frac{dy}{dx}=48y - 6x^{5}\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(6y^{5}-48x)=48y - 6x^{5}\).
Then \(\frac{dy}{dx}=\frac{48y - 6x^{5}}{6y^{5}-48x}\).
Simplify the fraction by factoring out a \(6\) from the numerator and the denominator: \(\frac{dy}{dx}=\frac{8y - x^{5}}{y^{5}-8x}\).

Answer:

\(\frac{8y - x^{5}}{y^{5}-8x}\)