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Differentiate y equals the quotient of the quantity 1 plus sine x and the quantity 1 minus cosine x.

–1

–2csc(x)sec(x)

2csc(x)sec(x)

the quotient of negative 1 times sine x plus cosine x minus 1 and the square of the quantity 1 minus cosine x

Differentiate y equals the quotient of the quantity 1 plus sine x and the quantity 1 minus cosine x 1 2cscxsecx 2cscxsecx the quotient of negative 1 times sine class=

Respuesta :

apologiabiology
trig was never my strong point but ok


remember the quotient rule
[tex]\frac{dy}{dx} \frac{f(x)}{g(x)}=\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}[/tex]

so
remember the pythagorean identity sin²(x)+cos²(x)=1
so

[tex]\frac{dy}{dx} \frac{1+sin(x)}{1-cos(x)}=\frac{cos(x)(1-cos(x))-sin(x)(1+sin(x))}{(1+cos(x))^2}=[/tex]
[tex]\frac{cos(x)-cos^2(x)-sin(x)-sin^2(x)}{(1+cos(x))^2}=[/tex]
[tex]\frac{cos(x)-sin(x)-sin^2(x)-cos^2(x)}{(1+cos(x))^2}=[/tex]
[tex]\frac{cos(x)-sin(x)-(sin^2(x)+cos^2(x))}{(1+cos(x))^2}=[/tex]
[tex]\frac{cos(x)-sin(x)-(1)}{(1+cos(x))^2}=[/tex]
[tex]\frac{cos(x)-sin(x)-1}{(1+cos(x))^2}=[/tex]
[tex]\frac{-sin(x)+cos(x)-1}{(1+cos(x))^2}=[/tex]

taht is the last option
thanks to jdoe0001 for showing me which identity to use

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