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