# Thread: Equation of tangent line

1. ## Equation of tangent line

Hey guys, hope I can get some help on this problem that I've been stuck on.

Find the equation of the tangent line to the given curve at the specific point.
y ln y = x, (0,1)

Thanks.

2. Originally Posted by drahcirnaw
Hey guys, hope I can get some help on this problem that I've been stuck on.

Find the equation of the tangent line to the given curve at the specific point.
y ln y = x, (0,1)

Thanks.
$\frac{d}{dx}(y\ln{y}) = \frac{d}{dx}(x)$

$\frac{d}{dy}(y\ln{y})\,\frac{dy}{dx} = 1$

$\left(y\frac{1}{y} + \ln{y}\right)\frac{dy}{dx} = 1$

$(1 + \ln{y})\frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{1 + \ln{y}}$.

So at the point $(0, 1)$

$\frac{dy}{dx} = \frac{1}{1 + \ln{1}}$

$= \frac{1}{1 + 0}$

$= 1$.

So now you have the gradient of the tangent and a point on the tangent line.

$y = mx + c$

$1 = 1(0) +c$

$c = 1$.

So the tangent line is $y = x + 1$.

3. Originally Posted by drahcirnaw
Hey guys, hope I can get some help on this problem that I've been stuck on.

Find the equation of the tangent line to the given curve at the specific point.
y ln y = x, (0,1)

Thanks.
Dear drahcirnaw,