# (Sinx-cosx)/(sinx+cosx) Show that (dy)/(dx)=1+y^2?

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#### Explanation

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#### Explanation:

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Jim G. Share
Jan 11, 2018

$\text{see explanation}$

#### Explanation:

$\text{let } y = \frac{\sin x - \cos x}{\sin x + \cos x}$

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "y=(g(x))/(h(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$

$g \left(x\right) = \sin x - \cos x \Rightarrow g ' \left(x\right) = \cos x + \sin x$

$h \left(x\right) = \sin x + \cos x \Rightarrow h ' \left(x\right) = \cos x - \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\sin x + \cos x\right) \left(\sin x + \cos x\right) - \left(\sin x - \cos x\right) - 1 \left(\sin x - \cos x\right)}{\sin x + \cos x} ^ 2$

$= \frac{{\left(\sin x + \cos x\right)}^{2} + {\left(\sin x - \cos x\right)}^{2}}{\sin x + \cos x} ^ 2$

$= {\left(\sin x + \cos x\right)}^{2} / {\left(\sin x + \cos x\right)}^{2} + {\left(\sin x - \cos x\right)}^{2} / {\left(\sin x + \cos x\right)}^{2}$

$= 1 + {y}^{2}$

Then teach the underlying concepts
Don't copy without citing sources
preview
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Jan 11, 2018

My explanation is below.

#### Explanation:

$y = \frac{\sin x - \cos x}{\sin x + \cos x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\cos x + \sin x\right) \left(\sin x + \cos x\right) - \left(\cos x - \sin x\right) \cdot \left(\sin x - \cos x\right)}{\sin x + \cos x} ^ 2$

=$\frac{{\left(\sin x + \cos x\right)}^{2} + {\left(\sin x - \cos x\right)}^{2}}{\sin x + \cos x} ^ 2$

=$1 + {\left[\frac{\sin x - \cos x}{\sin x + \cos x}\right]}^{2}$

=$1 + {y}^{2}$

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