How do you differentiate $y = \frac{1}{sin x} + \frac{1}{cos x}$ $y=1/sinx+1/cosx$ ?

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How do you differentiate $y = \frac{1}{sin x} + \frac{1}{cos x}$ $y=1/sinx+1/cosx$ ?

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Answer 1 · 2016-10-31T19:12:34

$\frac{d y}{d x} = - cot x csc x + tan x sec x$ $dy/dx = -cotx cscx + tanx secx$

Explanation:

You should learn that $\frac{1}{sin x} = csc x$ $1/sinx=cscx$ and $\frac{1}{cos x} = sec x$ $1/cosx=secx$
And, $\frac{d}{d x} csc x = - csc x cot x$ $d/dxcscx=-csc x cot x$ and $\frac{d}{d x} sec x = sec x tan x$ $d/dxsecx=sec x tan x$

These resul;ts can be obtained using the quotient rule;
$\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$ $d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2$ ;

$y = \frac{1}{sin x} + \frac{1}{cos x}$ $y = 1/sinx + 1/cosx$

$∴ \frac{d y}{d x} = ⎧ ⎪ ⎨ ⎪ ⎩ \frac{(sin x) (\frac{d}{d x} 1) - (1) (\frac{d}{d x} sin x)}{{(sin x)}^{2}} ⎫ ⎪ ⎬ ⎪ ⎭ + ⎧ ⎪ ⎨ ⎪ ⎩ \frac{(cos) (\frac{d}{d x} 1) - (1) (\frac{d}{d x} cos x)}{{(cos x)}^{2}} ⎫ ⎪ ⎬ ⎪ ⎭$ $:. dy/dx = { ( (sinx)(d/dx1) - (1)(d/dxsinx) ) / (sinx)^2 } + { ( (cos)(d/dx1) - (1)(d/dxcosx) ) / (cosx)^2 }$

$∴ \frac{d y}{d x} = {\frac{0 - (1) (cos x)}{{(sin x)}^{2}}} + {\frac{0 - (1) (- sin x)}{{(cos x)}^{2}}}$ $:. dy/dx = { ( 0 - (1)(cosx) ) / (sinx)^2 } + { ( 0 - (1)(-sinx) ) / (cosx)^2 }$

$∴ \frac{d y}{d x} = - \frac{cos x}{{(sin x)}^{2}} + \frac{sin x}{{(cos x)}^{2}}$ $:. dy/dx = -cosx/ (sinx)^2 + sinx / (cosx)^2$
$∴ \frac{d y}{d x} = - \frac{cos x}{sin x} \frac{1}{sin x} + \frac{sin x}{cos x} \frac{1}{cos x}$ $:. dy/dx = -cosx/sinx 1/sinx + sinx/cosx 1/cosx$
$∴ \frac{d y}{d x} = - cot x csc x + tan x sec x$ $:. dy/dx = -cotx cscx + tanx secx$

Answer 2 · 2016-10-31T19:26:29

$y = \frac{1}{sin x} + \frac{1}{cos x}$ $y=1/sinx+1/cosx$

$y = csc x + sec x$ $y=cscx+secx$

$\frac{d y}{d x} = - cot x csc x + sec x tan x$ $(dy)/(dx)=-cotxcscx+secxtanx$

To find out how to differentiate $csc x$ $cscx$ using implicit differentiation watch this video:

To find out how to differentiate $sec x$ $secx$ using implicit differentiation watch this video:

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