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
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