What is the derivative of y= 1/((sin(x)+cos(x))) ?

The derivative of y= 1/((sin(x)+cos(x))) is -(cos(x)-sin(x))/(1+sin(2x))

What is the first derivative of y= 1/((sin(x)+cos(x))) ?

The first derivative of y= 1/((sin(x)+cos(x))) is -(cos(x)-sin(x))/(1+sin(2x))

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derivative of

y = \frac{1}{( sin ( x ) + cos ( x ) )}

- \frac{cos ( x ) - sin ( x )}{1 + sin ( 2 x )}

Solution steps

\frac{d}{d x} (\frac{1}{sin ( x ) + cos ( x )})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((sin (x) + cos (x))^{- 1})

Apply the chain rule:

- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} \frac{d}{d x} (sin (x) + cos (x))

= - \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} \frac{d}{d x} (sin (x) + cos (x))

\frac{d}{d x} (sin (x) + cos (x)) = cos (x) - sin (x)

= - \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x))

Simplify

- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x)) : - \frac{cos ( x ) - sin ( x )}{1 + sin ( 2 x )}

= - \frac{cos ( x ) - sin ( x )}{1 + sin ( 2 x )}

cos (6 - 2 x)

t^{2} + 4 t + 5

x^{2} - cos (x) + 2

\frac{sin ( 1 )}{x}

\frac{d}{d x} (x^{2} \cdot y - 4 x^{3} + 3 \cdot 4^{2})

What is the derivative of y= 1/((sin(x)+cos(x))) ?
The derivative of y= 1/((sin(x)+cos(x))) is -(cos(x)-sin(x))/(1+sin(2x))
What is the first derivative of y= 1/((sin(x)+cos(x))) ?
The first derivative of y= 1/((sin(x)+cos(x))) is -(cos(x)-sin(x))/(1+sin(2x))