What is the derivative of $tan (x - y) = \frac{y}{1 + x^{2}}$ $tan(x-y)=y/(1+x^2)$ ?

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What is the derivative of $tan (x - y) = \frac{y}{1 + x^{2}}$ $tan(x-y)=y/(1+x^2)$ ?

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Answer 1 · 2018-05-07T14:31:25

$\frac{d y}{d x} = \frac{x^{4} + 2 x^{2} + y^{2} + 2 x y + 1}{x^{4} + 3 x^{2} + y^{2} + 2}$ $dy/dx = (x^4+2x^2 + y^2 + 2xy+1)/(x^4+3x^2 + y^2 +2)$

Explanation:

Differentiate both sides of the equality with respect to $x$ $x$ :

${sec}^{2} (x - y) (1 - \frac{d y}{d x}) = \frac{1}{1 + x^{2}} \frac{d y}{d x} - \frac{2 x y}{{(1 + x^{2})}^{2}}$ $sec^2(x-y) (1-dy/dx) = 1/(1+x^2)dy/dx - (2xy)/(1+x^2)^2$

solve for $\frac{d y}{d x}$ $dy/dx$ :

$\frac{d y}{d x} ({sec}^{2} (x - y) + \frac{1}{1 + x^{2}}) = {sec}^{2} (x - y) + \frac{2 x y}{{(1 + x^{2})}^{2}}$ $dy/dx (sec^2(x-y) +1/(1+x^2)) = sec^2(x-y) + (2xy)/(1+x^2)^2$

$\frac{d y}{d x} = \frac{{sec}^{2} (x - y) + \frac{2 x y}{{(1 + x^{2})}^{2}}}{{sec}^{2} (x - y) + \frac{1}{1 + x^{2}}}$ $dy/dx = (sec^2(x-y) + (2xy)/(1+x^2)^2)/(sec^2(x-y) +1/(1+x^2))$

Note now that:

${sec}^{2} (x - y) = 1 + {tan}^{2} (x - y) = 1 + \frac{y^{2}}{{(1 + x^{2})}^{2}}$ $sec^2(x-y) = 1+ tan^2(x-y) = 1+ y^2/(1+x^2)^2$

so:

$\frac{d y}{d x} = \frac{1 + \frac{y^{2}}{{(1 + x^{2})}^{2}} + \frac{2 x y}{{(1 + x^{2})}^{2}}}{1 + \frac{y^{2}}{{(1 + x^{2})}^{2}} + \frac{1}{1 + x^{2}}}$ $dy/dx = (1+ y^2/(1+x^2)^2 + (2xy)/(1+x^2)^2)/(1+ y^2/(1+x^2)^2 +1/(1+x^2))$

and multiplying numerator and denominator by ${(1 + x^{2})}^{2}$ $(1+x^2)^2$ :

$\frac{d y}{d x} = \frac{{(1 + x^{2})}^{2} + y^{2} + 2 x y}{{(1 + x^{2})}^{2} + y^{2} + 1 + x^{2}}$ $dy/dx = ((1+x^2)^2 + y^2 + 2xy)/((1+x^2)^2 + y^2 +1+x^2)$

$\frac{d y}{d x} = \frac{x^{4} + 2 x^{2} + y^{2} + 2 x y + 1}{x^{4} + 3 x^{2} + y^{2} + 2}$ $dy/dx = (x^4+2x^2 + y^2 + 2xy+1)/(x^4+3x^2 + y^2 +2)$

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What is the derivative of $tan (x - y) = \frac{y}{1 + x^{2}}$ $tan(x-y)=y/(1+x^2)$ ?

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