# determine the equation of the tangent line at x=1

#### nogoodmaths

It is given that e^(xy)=x[(x+1)^3] / (x^2+1) ,where x>0
determine the equation of the tangent line at x=1
my work::::::::xy ln e=lnx+3ln(x+1)-ln(x^2+1)

d/dy (xy ln e)=d/dy [(lnx+3ln(x+1)-ln(x^2+1)]
dy/dx(1/e)=1/x+[3(1/(x+1))]-[2x/(x^2+1)]
dy/dx =e*(1/x+[3(1/(x+1))]-[2x/(x^2+1)])
I know it is wrong and I don't know how to correct it and the next step?????

#### skeeter

MHF Helper
$e^{xy} = \dfrac{x(x+1)^3}{x^2+1}$

note that $\ln{e}=1$ ...

$xy = \ln{x} + 3\ln(x+1) - \ln(x^2+1)$

when $x=1$, $y=\ln{4}$

$\dfrac{d}{dx}\bigg[xy = \ln{x} + 3\ln(x+1) - \ln(x^2+1)\bigg]$

product rule on the left side of the equation ...

$x \cdot \dfrac{dy}{dx} + y = \dfrac{1}{x} + \dfrac{3}{x+1} - \dfrac{2x}{x^2+1}$

now evaluate the value of $\dfrac{dy}{dx}$ when $x=1$ and $y=\ln{4}$ to determine the slope, $m$, of the tangent line.

$y - \ln{4} = m(x - 1)$

1 person

#### nogoodmaths

Re: skeeter

So is the m=(1/2-ln4)?

$e^{xy} = \dfrac{x(x+1)^3}{x^2+1}$

note that $\ln{e}=1$ ...

$xy = \ln{x} + 3\ln(x+1) - \ln(x^2+1)$

when $x=1$, $y=\ln{4}$

$\dfrac{d}{dx}\bigg[xy = \ln{x} + 3\ln(x+1) - \ln(x^2+1)\bigg]$

product rule on the left side of the equation ...

$x \cdot \dfrac{dy}{dx} + y = \dfrac{1}{x} + \dfrac{3}{x+1} - \dfrac{2x}{x^2+1}$

now evaluate the value of $\dfrac{dy}{dx}$ when $x=1$ and $y=\ln{4}$ to determine the slope, $m$, of the tangent line.

$y - \ln{4} = m(x - 1)$

MHF Helper

#### nogoodmaths

Re: skeeter

is that 3/2-ln4?thx

#### HallsofIvy

MHF Helper
Personally, I would not have solved for y, taking logarithms to begin with. (Perhaps I just don't like logarithms!)

I would have first determined that, when x= 1, we have $$\displaystyle e^{y}= \frac{1(1+ 1)^3}{1^2+ 1}= \frac{8}{2}= 4$$ so that y= ln(4).

Then, using "implicit differentiation" $$\displaystyle e^{xy}+ e^{xy}y'= \frac{(x+1)^3}{x^2+ 1}+ \frac{3x(x+1)^2}{x^2+ 1}- \frac{2x^2(x+ 1)^3}{(x^2+ 1)^2}$$ so that, with x= 1, y= 4,
$$\displaystyle e^4+ e^4y'= \frac{(1+ 1)^3}{1+ 1}+ \frac{3(1)(1+1)^2}{1+ 1}- \frac{2(1)(1+1)^3}{(1+ 1)^2}$$
$$\displaystyle e^4(1+ y')= \frac{8}{2}+ \frac{3(4)}{2}- \frac{2(8)}{4}= 4+ 6- 4= 6$$
so that $$\displaystyle 1+ y'= 6e^{-4}$$ and $$\displaystyle y'= 6e^{-4}- 1$$.

The equation of the tangent line is $$\displaystyle y= (6e^{-4}-1)(x- 1)+ ln(4)$$.