Thread: Find the Equation of the Tangent Line to...

1. Find the Equation of the Tangent Line to...

Hello, and thanks in advance for the help.

The problem I'm given in this review for my upcoming exam is to find the equation of the tangent line to x = y - cos(y) at the point (-1, 0).

I've tried several things and am just hoping one of the local math gurus could make sure I've done my steps correctly :-).

I know I need to find y' at the point (-1, 0) because it's the slope of the tangent line. My first attempt at finding y' yielded y' = 1 / sin(y), where inputting 0 for y caused 1 / 0.

However, my second attempt is much more promising if it leads where I think, and I'll just list the steps below:

x = y - cos(y)
x - y = -cos(y)
1 - y'(y^0) = sin(y) [took the derivative of both sides, and y^0 = 1)
y' = 1 + sin(y)

Does this mean that at the point (-1, 0), the slope is 1 + sin(0) or m = 1?

If so, the equation for the tangent line would be y = x + 1, right?

Thanks again, this forum is a great resource!

2. Originally Posted by NBrunk

The problem I'm given in this review for my upcoming exam is to find the equation of the tangent line to x = y - cos(y) at the point (-1, 0).
you "stumbled" into the correct tangent line equation because the slope is 1 at (-1,0)

here is the correct derivative ...

$\displaystyle \frac{d}{dx}(x = y - \cos{y})$

$\displaystyle 1 = y' + sin{y} \cdot y'$

$\displaystyle 1 = y'(1 + \sin{y})$

$\displaystyle \frac{1}{1+\sin{y}} = y'$

3. Funny how that worked out, and careless of me to forget to add y' to the cos(y) term for the derivative.

Thank you, I understand this problem now.