Another differential equation help...

**omibayne** · October 26th 2008, 08:32 PM

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

**Peritus** · October 27th 2008, 02:51 AM

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

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

$ \Leftrightarrow y' = \frac{{\tan \left( {20y^{12} } \right)}} {{240y^{11} }} $

**toraj58** · October 27th 2008, 03:30 AM

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 $\frac {1}{\tan}$

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