Question

If xy+6e^y = 6e, find the value of y\" at the point where x = 0.

Answer 1

Xy + 6e^y = 6e
taking first derivative : 
xy' + y + 6y(e^y)(y') = 0 
xy" + y' + y' + 6y'(e^y)(y') + 6yy(e^y)y' + 6y(e^y)y" = 0 

Put x = 0,
0 + 6e^y = 6e,
y = 1 
0 + 1 + 6(1)(y')e = 0,
Put y' = 0 
0 + 0 + 0 + 0 + 0 +6 ey" = 0,
y" = 0

Answer 2

The value of $y^{"}$ at the point at $x=0$ is $\boxed{\dfrac{y(2-y)}{36e^{2y}}}$.

Further explanation:

The given equation is $xy+6e^{y}=6e$.

Now we have to find $y^{"}$ that means differentiate the given equation $xy+6e^{y}=6e$ twice with respect to $x$ to obtain the second derivative of the equation.

Differentiation is a technique to find out the derivative of an equation.

The given equation is in product form, so we will use product rule of differentiation of a function.

If $u$ and $v$ are two functions in product form then the derivative of this product form is given as  follows:

$\boxed{\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}}$            .....(1)

Differentiate the equation $xy+6e^{y}=6e$ with respect to $x$ on both side to obtain the first derivative by the use of product rule.

$\begin{aligned}\dfrac{d}{dx}(xy+6e^{y})&=0\\ \dfrac{d}{dx}(xy)+\dfrac{d}{dx}(6e^{y})&=0\\x\dfrac{dy}{dx}+y+6e^{y}\dfrac{dy}{dx}&=0\\(x+6e^{y})\dfrac{dy}{dx}&=-y\\ \dfrac{dy}{dx}&=-\dfrac{y}{x+6e^{y}}\end{aligned}$  

So the first derivative of the given equation is calculated above.  

Now to find the second derivative we will use the division rule of the differentiation because the first derivative that we have calculated is in division form.

If $u$ and $v$ are two functions in division form then the derivative of this division form is given as follows:

$\boxed{\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}}}$               .......(2)

Now differentiate the first derivative with respect to $x$ by the use of division rule to obtain the second derivative as follows,

$\begin{aligned}\dfrac{d^{2}y}{dx^{2}}&=-\dfrac{d}{dx}\left(\dfrac{y}{x+6e^{y}}\right)\\&=-\left(\dfrac{(x+6e^{y})\frac{dy}{dx}-y\frac{d}{dx}(x+6e^{y})}{(x+6e^{y})^{2}}\right)\\&=-\left(\dfrac{(x+6e^{y}-6ye^{y})\frac{dy}{dx}-y}{(x+6e^{y})^{2}}\right)\end{aligned}$  

Now substitute the value of first derivative in the above equation to simplify the equation further,

$\begin{aligned}\dfrac{d^{2}y}{dx^{2}}&=-\left(\dfrac{(x+6e^{y}-6ye^{y})(\frac{-y}{x+6e^{y}})-y}{(x+6e^{y})^{2}}\right)\\&=\dfrac{(x+6e^{y}-6ye^{y})(\frac{y}{x+6e^{y}})+y}{(x+6e^{y})^{2}}\\&=\dfrac{(xy+6ye^{y}-6y^{2}e^{y})+(xy+6ye^{y})}{(x+6e^{y})^{3}}\\&=\dfrac{2xy+12ye^{y}-6y^{2}e^{y}}{(x+6e^{y})^{3}}\end{aligned}$  

The above result is the second derivative of the equation $xy+6e^{y}=6e$.

Substitute $x=0$ in the above result to obtain the second derivative at $x=0$.

$\begin{aligned}\dfrac{d^{2}y}{dx^{2}}&=\dfrac{(0+12ye^{y}-6y^{2}e^{y})}{(0+6e^{y})^{3}}\\&=\dfrac{12ye^{y}-6y^{2}e^{y}}{(6e^{y})^{3}}\\&=\dfrac{6e^{y}(2y-y^{2})}{6e^{y}(6e^{2y})}\end{aligned}$  

Therefore the second derivative of equation $xy+6e^{y}=6e$ at $x=0$ is $\boxed{\dfrac{y(2-y)}{36e^{2y}}}$.

Learn more:

1. Problem on equation of a line in slope intercept form: brainly.com/question/1473992

2. Problem on the range of a function: brainly.com/question/1435353

3. Problem on inverse of a function: brainly.com/question/1632445

Answer detail:

Grade: College

Subject: Mathematics

Chapter: Differential calculaus

Keywords: Differentiation, function, equation, xy+6e^y=6e , first derivative, second derivative, simplify, product rule, division rule, integration, calculus, order, y(2-y)/36e^2y.