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.
$\displaystyle \frac{d}{dx}(y\ln{y}) = \frac{d}{dx}(x)$
$\displaystyle \frac{d}{dy}(y\ln{y})\,\frac{dy}{dx} = 1$
$\displaystyle \left(y\frac{1}{y} + \ln{y}\right)\frac{dy}{dx} = 1$
$\displaystyle (1 + \ln{y})\frac{dy}{dx} = 1$
$\displaystyle \frac{dy}{dx} = \frac{1}{1 + \ln{y}}$.
So at the point $\displaystyle (0, 1)$
$\displaystyle \frac{dy}{dx} = \frac{1}{1 + \ln{1}}$
$\displaystyle = \frac{1}{1 + 0}$
$\displaystyle = 1$.
So now you have the gradient of the tangent and a point on the tangent line.
$\displaystyle y = mx + c$
$\displaystyle 1 = 1(0) +c$
$\displaystyle c = 1$.
So the tangent line is $\displaystyle y = x + 1$.