Question #faa9f

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Question #faa9f

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Answer 1 · 2017-10-05T05:16:17

$\frac{d y}{d x} = \frac{({sec}^{2} (x - y)) {(3 + x^{2})}^{2} + (y) (2 x)}{(3 + x^{2}) + ({sec}^{2} (x - y)) {(3 + x^{2})}^{2}}$ $dy/dx=((sec^2(x-y))(3+x^2)^2+(y)(2x))/((3+x^2)+(sec^2(x-y))(3+x^2)^2)$

Explanation:

$tan (x - y) = \frac{y}{3 + x^{2}}$ $tan(x-y)=y/(3+x^2)$

Take the derivative of both sides.

$\frac{d}{d x} (tan (x - y)) = \frac{d}{d x} (\frac{y}{3 + x^{2}})$ $d/dx(tan(x-y))=d/dx(y/(3+x^2))$

Solve.

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

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

Now we know:

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

We can simplify

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

$(({sec}^{2} (x - y)) {(3 + x^{2})}^{2} - (\frac{d y}{d x}) ({sec}^{2} (x - y)) {(3 + x^{2})}^{2}) = ((3 + x^{2}) (\frac{d y}{d x}) - (y) (2 x))$ $((sec^2(x-y))(3+x^2)^2-(dy/dx)(sec^2(x-y))(3+x^2)^2)=((3+x^2)(dy/dx)-(y)(2x))$

$(({sec}^{2} (x - y)) {(3 + x^{2})}^{2} + (y) (2 x)) = ((3 + x^{2}) (\frac{d y}{d x}) + (\frac{d y}{d x}) ({sec}^{2} (x - y)) {(3 + x^{2})}^{2})$ $((sec^2(x-y))(3+x^2)^2+(y)(2x))=((3+x^2)(dy/dx)+(dy/dx)(sec^2(x-y))(3+x^2)^2)$

$(({sec}^{2} (x - y)) {(3 + x^{2})}^{2} + (y) (2 x)) = (\frac{d y}{d x}) ((3 + x^{2}) + ({sec}^{2} (x - y)) {(3 + x^{2})}^{2})$ $((sec^2(x-y))(3+x^2)^2+(y)(2x))=(dy/dx)((3+x^2)+(sec^2(x-y))(3+x^2)^2)$

$\frac{({sec}^{2} (x - y)) {(3 + x^{2})}^{2} + (y) (2 x)}{(3 + x^{2}) + ({sec}^{2} (x - y)) {(3 + x^{2})}^{2}} = \frac{d y}{d x}$ $((sec^2(x-y))(3+x^2)^2+(y)(2x))/((3+x^2)+(sec^2(x-y))(3+x^2)^2)=dy/dx$

Depending on how simplified the answer needs to be, this is technically the solution.

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Question #faa9f

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