# using implicit differentiation and a point to find equation of the tangent line.

• Feb 28th 2012, 02:22 PM
escreality
using implicit differentiation and a point to find equation of the tangent line.
I've been struggling to work on this problem.

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
I know I differentiate the equation to get slope of the tangent, but I have no idea what to do afterwards, or frankly if my differentiation is even correct.

y sin 12x = x cos 2y, (pi/2, pi/4)

Any help is greatly appreciated.
• Feb 28th 2012, 03:27 PM
skeeter
Re: using implicit differentiation and a point to find equation of the tangent line.
Quote:

Originally Posted by escreality
I've been struggling to work on this problem.

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
I know I differentiate the equation to get slope of the tangent, but I have no idea what to do afterwards, or frankly if my differentiation is even correct.

y sin 12x = x cos 2y, (pi/2, pi/4)

Any help is greatly appreciated.

$\displaystyle \frac{d}{dx} [y\sin(12x) = x\cos(2y)]$

product rule, both sides ...

$\displaystyle y \cdot 12\cos(12x) + y' \cdot \sin(12x) = -x\sin(2y) \cdot 2y' + \cos(2y)$

sub in $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle y = \frac{\pi}{4}$ and solve for the value of $\displaystyle y'$ , the slope of the tangent line.