X*3 - 2x*2 y*2 + 5x + y - 5 = 0Then dy/dx at x = 1, y = 1 is equal to.

1.	X3 - 2x2 y*2 + 5x + y - 5 = 0Then dy/dx at x = 1, y = 1 is equal to..?
Answer» Given : Function $\bf\:x{}^{3}-2x{}^{2}\times\:y{}^{2}+5x+y-5=0$ To Find : $\dfrac{dy}{dx}$ at x=1 and y=1 Formula's: $1) \: \frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$ $2) \frac{d(constant)}{dx} = 0$ Solution : $\bf\:x{}^{3}-2x{}^{2}\times\:y{}^{2}+5x+y-5=0$ DIFFERENTIATE with respect to x $3x {}^{2} - (2x {}^{2} \times 2y \frac{dy}{dx} + y {}^{2} \times 4x) + 5 + \frac{dy}{dx} = 0$ $3x {}^{2} - 4x {}^{2} y \frac{dy}{dx} - 4x y {}^{2} + 5 + \frac{dy}{dx} = 0$ $3x {}^{2} - 4xy {}^{2} + 5 = \frac{dy}{dx} (4x {}^{2} y) - \frac{dy}{dx}$ $3x {}^{2} - 4xy {}^{2} + 5 = \frac{dy}{dx}(4xy {}^{2} - 1)$ $\implies \dfrac{dy}{dx} = \dfrac{3x {}^{2} - 4xy {}^{2} + 5 }{4x {}^{2} y - 1}$ $\implies \dfrac{dy}{dx} at (1,1) = \frac{3 - 4 + 5}{4 - 1}$ $\implies \bf \dfrac{dy}{dx} at (1,1)= \dfrac{4}{3}$ which is the required solution! _______________________________ More About Differention: • CHAIN rule Let y=f(t) ,t = g(u) and u =m(x) ,then $\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$