Equation of tangent line

**drahcirnaw** · March 11th 2010, 10:55 PM

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.

**Prove It** · March 11th 2010, 11:09 PM

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 $y = x + 1$ .

**Sudharaka** · March 11th 2010, 11:16 PM

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,

When I had written my answer Prove it had already posted his one. Therefore I deleted it.

**drahcirnaw** · March 11th 2010, 11:52 PM

Thanks for the help. Cleared up my problems.

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