What is the derivative of $\frac{d}{d x} (x + cos x \div tan x)$ $d/dx (x+cosx ÷ tanx )$ ?

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What is the derivative of $\frac{d}{d x} (x + cos x \div tan x)$ $d/dx (x+cosx ÷ tanx )$ ?

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Answer 1 · 2017-02-25T21:19:47

$\frac{d}{d x} (x + cos x \div tan x) = 1 - cos x - cot x csc x$ $d/dx (x+cosx ÷ tanx ) = 1 - cosx - cotxcscx$

Explanation:

$\frac{d}{d x} (x + cos x \div tan x) = \frac{d}{d x} (x + \frac{cos x}{tan x})$ $d/dx (x+cosx ÷ tanx ) = d/dx (x+cosx/tanx )$
$= \frac{d}{d x} (x) + \frac{d}{d x} (\frac{cos x}{tan x})$ $" "= d/dx (x)+d/dx(cosx/tanx)$
$= 1 + \frac{(tan x) (- sin x) - (cos x) ({sec}^{2} x)}{{(tan x)}^{2}}$ $" "= 1 + { (tanx)(-sinx) - (cosx)(sec^2x) }/ (tanx)^2$
$= 1 - (tan x) \frac{sin x}{{tan}^{2} x} - (cos x) \frac{{sec}^{2} x}{{(tan x)}^{2}}$ $" "= 1 - (tanx)(sinx)/tan^2x - (cosx)(sec^2x)/(tanx)^2$
$= 1 - (sin x) \cdot \frac{cos x}{sin x} - (cos x) (\frac{1}{{cos}^{2} x}) \cdot \frac{{cos}^{2} x}{{sin}^{2} x}$ $" "= 1 - (sinx)*cosx/sinx - (cosx)(1/cos^2x)*cos^2x/sin^2x$

$= 1 - cos x - \frac{cos x}{{sin}^{2} x}$ $" "= 1 - cosx - (cosx)/sin^2x$

$= 1 - cos x - cot x csc x$ $" "= 1 - cosx - cotxcscx$

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What is the derivative of $\frac{d}{d x} (x + cos x \div tan x)$ $d/dx (x+cosx ÷ tanx )$ ?

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Explanation:

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