What is the derivative of 8/(sin(x)+cos(x)) ?

The derivative of 8/(sin(x)+cos(x)) is -(8(cos(x)-sin(x)))/(1+sin(2x))

What is the first derivative of 8/(sin(x)+cos(x)) ?

The first derivative of 8/(sin(x)+cos(x)) is -(8(cos(x)-sin(x)))/(1+sin(2x))

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derivative of

\frac{8}{sin ( x ) + cos ( x )}

- \frac{8 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

Solution steps

\frac{d}{d x} (\frac{8}{sin ( x ) + cos ( x )})

Take the constant out:

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

= 8 \frac{d}{d x} (\frac{1}{sin ( x ) + cos ( x )})

Apply exponent rule:

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

= 8 \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)

= 8 (- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x)))

Simplify

8 (- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x))) : - \frac{8 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

= - \frac{8 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

f (x) = 4 x^{2} - 5 x (- 1.9)

\int_{3}^{4} \frac{1}{( x - 3 ) ^{2}} d x

(\frac{3}{5}, \frac{2}{3}), (- \frac{3}{5}, \frac{3}{4})

\int (e^{t})^{2} d t

\int x^{3} sec^{2} (y) d y

What is the derivative of 8/(sin(x)+cos(x)) ?
The derivative of 8/(sin(x)+cos(x)) is -(8(cos(x)-sin(x)))/(1+sin(2x))
What is the first derivative of 8/(sin(x)+cos(x)) ?
The first derivative of 8/(sin(x)+cos(x)) is -(8(cos(x)-sin(x)))/(1+sin(2x))