Solution assignment 05 Implicit differentiation

Calculate the tangent line at the point $(0,0)$ of the following graph:

$(x+1)e^y-1=0$

First we have to verify that the point $(0,0)$ lies on the graph, and yes, we have:

$(0+1)e^0-1=0$

Next we have to determine the derivative $y'$ and use both the product and quotient rule:

$1\cdot{e^y}+(x+1)e^yy'=0$

and thus:

$y'=-\displaystyle\frac{1}{x+1}$

and thus in the point $(0,0)$ :

$y'=-1$

The equation of the tangent line becomes:

$(y-0)=-1(x-0)$ or $y=-x$ .

Also in this case implicit differentiation is not necessary. We can write the function $y$ in explicit form quite easily:

$e^y=\displaystyle\frac{1}{x+1}$

and thus, after applying the logarithm in the left- and right-hand side we get:

$y=\ln(\displaystyle\frac{1}{x+1})=-\ln(x+1)$

Differentiation of this function gives the result found above.