How do you use the limit definition to find the derivative of $y = csc x$ $y = cscx$ ?

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How do you use the limit definition to find the derivative of $y = csc x$ $y = cscx$ ?

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Answer 1 · 2016-12-02T14:39:06

The key parts are in the explanation section, below.

Explanation:

$\frac{\frac{1}{sin (x + h)} - \frac{1}{sin x}}{h} = \frac{sin x - sin (x + h)}{h sin (x + h) sin x}$ $(1/sin(x+h)-1/sinx)/h = (sinx-sin(x+h))/(hsin(x+h)sinx)$

$= \frac{sin x - sin x cos h - cos x sin h}{h sin (x + h) sin x}$ $= (sinx-sinxcos h-cosxsin h)/(hsin(x+h)sinx)$

$= (sin x \frac{1 - cos h}{h} - cos x \frac{sin h}{h}) \cdot \frac{1}{sin (x + h) sin x}$ $= (sinx(1-cos h)/h - cosxsin h/h)*1/(sin(x+h)sinx)$

Taking the limit gets us

$\frac{- cos x}{{sin}^{2} x} = - csc x cot x$ $(-cosx)/sin^2 x = -cscxcotx$

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How do you use the limit definition to find the derivative of $y = csc x$ $y = cscx$ ?

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Explanation:

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How do you use the limit definition to find the derivative of $y = csc x$ $y = cscx$ ?