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

This question aims to find the second derivative of the given implicit function. A function’s derivatives describe the rate of change of that function at a given point.

If the dependent variable, say $y$, is a function of the independent variable, say $x$, we usually express $y$ in terms of $x$. When this occurs, $y$ is said to be an explicit function of $x$.

For instance, when we express $y=x^2+2x$, this means we are defining $y$ explicitly in terms of $x$. If the relation between the values $y$ and $x$ is depicted by an equation where $y$ is not completely stated in terms of $x$, the equation is said to implicitly define $y$ in terms of $x$. The equation $\cos(y)+y=x^2+3$ is an example of an implicit equation.

We can use implicit differentiation to find slopes of tangents to curves that are explicitly not functions. This means that some components of $y$ are the functions that satisfy the given equation, but $y$ itself is not a function of $x$. The Chain Rule-based technique of implicit differentiation is used to find a derivative in the case when the relation between the variables is expressed implicitly rather than explicitly.

Expert Answer

The given equation is:

$xy+6e^y=6e$                             $(1)$

Put $x=0$ in $(1)$


$\implies 6e^y=6e\implies e^y=e$

$\implies y=1$

Hence, we have $y=1$ for $x=0$.

Now, differentiating both sides of $(1)$ with respect to $x$, we get:

$xy’+y+6e^yy’=0$                                 $(2)$

Putting  $x=0$ and $y=1$ in  $(2)$, we obtain:


$\implies 1+6ey’=0$

$\implies y’=\dfrac{-1}{6e}$

Differentiating both sides of $(2)$ again with respect to $x$, we get:


$\implies xy”+6e^yy”+2y’+6e^y(y’)^2=0$                          $(3)$

Plugging the values of  $x,y$  and $y’$ in $(3)$, we get


$\implies 6ey”-\dfrac{1}{3e}+\dfrac{1}{6e}=0$

$\implies 6ey”-\dfrac{1}{6e}=0$

$\implies 6ey”=\dfrac{1}{6e}$

$\implies y”=\dfrac{1}{36e^2}$

Graph of the given implicit equation:


Find $y”$ when $x^2+y^2=4$.


Differentiating the given equation with respect to $x$, we get:


$\implies y’=-\dfrac{x}{y}$                           $(1)$

Differentiating $(1)$ again with respect to $x$, we get:


$\implies y”=-\dfrac{y-xy’}{y^2}$                        $(2)$

Substituting $(1)$ in $(2)$


$\implies y”=-\dfrac{y^2+x^2}{y^3}$

Images/mathematical drawings are created with GeoGebra.

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