What is the derivative of 2/(sin(x+cos(x))) ?

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

What is the first derivative of 2/(sin(x+cos(x))) ?

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

= 2 \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)

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

Simplify

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

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

t \frac{d y}{d t} = t^{3} + 25 t^{3} y

\int (y sin (x y)) d x

\frac{\partial}{\partial x} (y^{2} - 4 x^{2} + 7 y + 5 x + 13)

y^{^{'}} + 5 (cos (5 x)) y = - 8 cos (5 x)

y^{2} = 4 x, y^{2} = 8 x - 4, x

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