# Calculus - Implicit differentiation

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• Mar 27th 2008, 03:59 AM
ssdimensionss
Calculus - Implicit differentiation
consider: e^(xy)-ycosh(4x)=-sinx

use implicit differentiation to find dy/dx and then determine the equation of the tangent to the curve at the point (0,1)

i did the implicit differentiation but it turned out to b really messy plz help! thx
• Mar 27th 2008, 04:17 AM
topsquark
Quote:

Originally Posted by ssdimensionss
consider: e^(xy)-ycosh(4x)=-sinx

use implicit differentiation to find dy/dx and then determine the equation of the tangent to the curve at the point (0,1)

i did the implicit differentiation but it turned out to b really messy plz help! thx

$\displaystyle e^{xy} - y~cosh(4x) = -sinx$

$\displaystyle ye^{xy} + e^{xy} \cdot xy^{\prime} - y^{\prime}~cosh(4x) - 4y~sinh(4x) = -cos(x)$

$\displaystyle (xe^{xy} - cosh(4x)) y^{\prime} = 4y~sinh(4x) - cos(x) - ye^{xy}$

$\displaystyle y^{\prime} = \frac{4y~sinh(4x) - cos(x) - ye^{xy}}{xe^{xy} - cosh(4x)}$

At the point (0, 1) we get
$\displaystyle y^{\prime} = \frac{4sinh(0) - cos(0) - e^{0}}{-cosh(0)} = 2$

So what is the line with a slope of 2 that passes through the point (0, 1)?

-Dan