Question

x²+3xy + y² = 11 dy upon dx at (1,1)​

Answer 1

Topic :-

Differentiation

Given :-

x² + 3xy + y² = 11

To Find :-

$\sf {\dfrac{dy}{dx}\:\:at\:\:(1,1)}$

Solution :-

$\sf {x^2+3xy+y^2=11}$

$\sf {\dfrac{d(x^2+3xy+y^2)}{dx}=\dfrac{d(11)}{dx}}$

From Sum Rule, we can write,

$\sf {\dfrac{d(x^2)}{dx}+\dfrac{d(3xy)}{dx}+\dfrac{d(y^2)}{dx}=\dfrac{d(11)}{dx}}$

$\sf {\dfrac{d(x^2)}{dx}+3\dfrac{d(xy)}{dx}+\dfrac{d(y^2)}{dx}=\dfrac{d(11)}{dx}}$

$\sf {\left( \because \dfrac{d(k.f(x))}{dx}=k\dfrac{d(f(x))}{dx},where\:k\:is\:a\:constant.\right)}$

$\sf {2x+3\dfrac{d(xy)}{dx}+\dfrac{d(y^2)}{dx}=\dfrac{d(11)}{dx}}$

$\sf {\left( \because \dfrac{d(x^n)}{dx}=nx^{n-1}\right)}$

$\sf {2x+3\left(y\dfrac{dx}{dx}+x\dfrac{dy}{dx}\right)+\dfrac{d(y^2)}{dx}=\dfrac{d(11)}{dx}}$

$\sf {\left(\because \dfrac{d(fg)}{dx}=g\dfrac{df}{dx}+f\dfrac{dg}{dx}\right)}$

$\sf {2x+3\left(y+x\dfrac{dy}{dx}\right)+\dfrac{d(y^2)}{dx}=\dfrac{d(11)}{dx}}$

$\sf {\left( \because \dfrac{dx}{dx}=1\right)}$

$\sf {2x+3y+3x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=\dfrac{d(11)}{dx}}$

$\sf {\left( \because \dfrac{d(y^n)}{dx}=ny^{n-1}.\dfrac{dy}{dx}\right)}$

$\sf {2x+3y+\left(3x+2y\right)\dfrac{dy}{dx}=0}$

$\sf {\left( \because \dfrac{dk}{dx}=0,where\:k\:is\:a\:constant.\right)}$

$\sf {\left(3x+2y\right)\dfrac{dy}{dx}=-(2x+3y)}$

$\sf {\dfrac{dy}{dx}=-\left(\dfrac{2x+3y}{3x+2y}\right)}$

Now,

x = 1

y = 1

Substituting Values,

$\sf {\dfrac{dy}{dx}=-\left(\dfrac{2(1)+3(1)}{3(1)+2(1)}\right)}$

$\sf {\dfrac{dy}{dx}=-\left(\dfrac{5}{5}\right)}$

$\sf {\dfrac{dy}{dx}=-1}$

Answer :-

$\sf {\dfrac{dy}{dx}\:\:at\:\:(1,1)=\bold{-1}}$