What is the derivative of this function $y = \frac{1 + sin x}{1 - sin x}$ $y= (1+sin x)/(1-sin x)$ ?

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What is the derivative of this function $y = \frac{1 + sin x}{1 - sin x}$ $y= (1+sin x)/(1-sin x)$ ?

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Answer 1 · 2016-11-05T13:36:04

$\frac{d y}{d x} = \frac{2 cos x}{{(1 - sin x)}^{2}}$ $dy/dx = ( 2cosx ) / (1-sinx)^2$

Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

$\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$ $d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2$ , or less formally, $(u/v)' = (v(du)-u(dv))/v^2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $y=(1+sinx)/(1-sinx)$ Then

${ ("Let "u=1+sinx, => , (du)/dx=cosx), ("And "v=1-sinx, =>, (dv)/dx=-cosx ) :}$

$:. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2$
$:. dy/dx = ( (1-sinx)(cosx) - (1+sinx)(-cosx) ) / (1-sinx)^2$
$:. dy/dx = ( cosx-sinxcosx + cosx +sinxcosx ) / (1-sinx)^2$
$:. dy/dx = ( 2cosx ) / (1-sinx)^2$

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What is the derivative of this function $y = \frac{1 + sin x}{1 - sin x}$ $y= (1+sin x)/(1-sin x)$ ?

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What is the derivative of this function y= (1+sin x)/(1-sin x)y=1+sinx1−sinxy= (1+sin x)/(1-sin x)?

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What is the derivative of this function $y = \frac{1 + sin x}{1 - sin x}$ $y= (1+sin x)/(1-sin x)$ ?