What is the derivative of $(\frac{12}{sin x}) + (\frac{1}{cot x})$ $(12/sinx) + (1/cotx)$ ?

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What is the derivative of $(\frac{12}{sin x}) + (\frac{1}{cot x})$ $(12/sinx) + (1/cotx)$ ?

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Answer 1 · 2015-09-16T14:31:59

$- 12 cos e c x \cdot cot x + {sec}^{2} x$ $−12cosecx⋅cotx+sec^2x$

Explanation:

1/ $sin x$ $sin x$ = $cos e c x$ $cosec x$ similarly $\frac{1}{cot x}$ $1/cotx$ = $tan x$ $tanx$
the question changes to $12 cos e c x + tan x$ $12cosecx+tanx$
the derivative is $- 12 cos e c x \cdot cot x + {sec}^{2} x$ $-12cosecx*cotx+sec^2x$ by (u/v rule of differentiation)
$\frac{d}{d x}$ $d/dx$ $(\frac{1}{sin x})$ $(1/sinx)$ = $\frac{1}{{sin}^{2} x}$ $1/sin^2x$ $(- 1 \cdot cos x)$ $(-1*cosx)$
= $- \frac{cos x}{sin x}$ $-cosx/sinx$ $\frac{1}{sin x}$ $1/sinx$
= $- cos e c x cot x$ $-cosecx cotx$
$\frac{d}{d x}$ $d/dx$ $tan x$ $tanx$ = $\frac{d}{d x} (\frac{sin x}{cos x})$ $d/dx(sinx/cosx)$
= $\frac{1}{{cos}^{2} x}$ $1/cos^2x$ $(cos x \cdot cos x - sin x \cdot (- sin x))$ $(cosx*cosx-sinx*(-sinx))$
= $\frac{{cos}^{2} x + {sin}^{2} x}{{cos}^{2} x}$ $(cos^2x+sin^2x)/cos^2x$
= $\frac{1}{{cos}^{2} x}$ $1/cos^2x$ = ${sec}^{2} x$ $sec^2x$

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What is the derivative of $(\frac{12}{sin x}) + (\frac{1}{cot x})$ $(12/sinx) + (1/cotx)$ ?

1 Answer

Explanation:

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What is the derivative of $(\frac{12}{sin x}) + (\frac{1}{cot x})$ $(12/sinx) + (1/cotx)$ ?