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How do you use implicit differentiation to find dy/dx given $x sin y + x^{2} cos y = 1$ $xsiny+x^2cosy=1$ ?

1 Answer

Oct 31, 2016

$\frac{d y}{d x} = - \frac{sin y + 2 x cos y}{(cos y) x - (sin y) x^{2}}$ $(dy)/(dx)=-(siny+2xcosy)/((cosy)x-(siny)x^2)$

Explanation:

Differentiation of the expression is determined by using the product rule differentiation

$\frac{d}{d x} (u (x) + v (x)) = \frac{d}{d x} (u (x)) + \frac{d}{d x} (v (x))$ $color(brown)(d/dx(u(x)+v(x))=d/dx(u(x))+d/dx(v(x)))$

$\frac{d}{d x} (u (x) \times v (x))$ $color(red)(d/dx(u(x)xxv(x))$
$= \frac{d u (x)}{d x} \times v (x) + \frac{d v (x)}{d x} \times u (x)$ $color(red)(=(du(x))/dxxxv(x)+(dv(x))/dxxxu(x)$

Applying the differentiation of trigonometric functions
$\frac{d}{d x} (sin x) = cos x$ $color(blue)(d/dx(sinx)=cosx)$
$\frac{d}{d x} (cos x) = - sin x$ $color(blue)(d/dx(cosx)=-sinx)$

$\frac{d}{d x} (x sin y + x^{2} cos y) = \frac{d 1}{d x}$ $d/dx(xsiny+x^2cosy)=(d1)/dx$

$\Rightarrow \frac{d}{d x} (x sin y) + \frac{d}{d x} (x^{2} cos y) = 0$ $rArrcolor(brown)(d/dx(xsiny)+d/dx(x^2cosy)=0)$

$\Rightarrow (\frac{d x}{d x} \times sin y + \frac{d sin y}{d x} \times x) + (\frac{{d x}^{2}}{d x} \times cos y + \frac{d cos y}{d x} \times x^{2}) = 0$ $rArrcolor(red)(((dx)/dxxxsiny+(dsiny)/dxxxx)+((dx^2)/dxxxcosy+(dcosy)/dxxxx^2)=0$

$\Rightarrow sin y + (\frac{d y}{d x}) (cos y) x + 2 x cos y - \frac{d y}{d x} (sin y) x^{2} = 0$ $rArrsiny+(dy/dx)(cosy)x+2xcosy-(dy)/dx(siny)x^2=0$

$\Rightarrow (\frac{d y}{d x}) (cos y) x - \frac{d y}{d x} (sin y) x^{2} = - sin y - 2 x cos y$ $rArr(dy/dx)(cosy)x-(dy)/dx(siny)x^2=-siny-2xcosy$

$\Rightarrow (\frac{d y}{d x}) ((cos y) x - (sin y) x^{2}) = - sin y - 2 x cos y$ $rArr(dy/dx)((cosy)x-(siny)x^2)=-siny-2xcosy$

$\Rightarrow \frac{d y}{d x} = - \frac{sin y + 2 x cos y}{(cos y) x - (sin y) x^{2}}$ $rArr(dy)/(dx)=-(siny+2xcosy)/((cosy)x-(siny)x^2)$

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