# Another differential equation help...

• Oct 26th 2008, 09:32 PM
omibayne
Another differential equation help...
Assume that the relation $x+20y^{12}+cot(20y^{12})=1$ defines y implicitly as a differentiable function of x. Find a simplified expression for $dy/dx$ in terms of the tangent function.

Any help will be much appreciated..thanks
• Oct 27th 2008, 03:51 AM
Peritus
$\begin{gathered}
\frac{d}
{{dx}}\left( {x + 20y^{12} + \cot \left( {20y^{12} } \right) = 1} \right) \hfill \\
\Leftrightarrow 1 + 240y^{11} y' - \frac{{240y^{11} y'}}
{{\sin ^2 \left( {20y^{12} } \right)}} = 0 \hfill \\
\end{gathered}$

$\begin{gathered}
\Leftrightarrow 1 + 240y^{11} y'\left( {1 - \frac{1}
{{\sin ^2 \left( {20y^{12} } \right)}}} \right) = 0 \hfill \\
\Leftrightarrow 1 - 240y^{11} y'\cot \left( {20y^{12} } \right) = 0 \hfill \\
\end{gathered}$

$
\Leftrightarrow y' = \frac{{\tan \left( {20y^{12} } \right)}}
{{240y^{11} }}
$
• Oct 27th 2008, 04:30 AM
toraj58
firsr sort it like this:

$x+20y^{12}+cot(20y^{12})-1=0$

then differntiate using this approach:

if $f(x,y) = 0$

then

$\frac {dy}{dx} = -\frac {f^\prime_x}{f^\prime_y}$

that means when you differntiate in respect to x then y is assumed as constant and in reverse.

then replace $\cot$ with $\frac {1}{\tan}$