What is the derivative of 9/(sin(x+cos(x))) ?

The derivative of 9/(sin(x+cos(x))) is -(9(cos(x)-sin(x)))/(1+sin(2x))

What is the first derivative of 9/(sin(x+cos(x))) ?

The first derivative of 9/(sin(x+cos(x))) is -(9(cos(x)-sin(x)))/(1+sin(2x))

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\frac{d}{d x} (\frac{9}{sin ( x ) + cos ( x )})

- \frac{9 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

Solution steps

\frac{d}{d x} (\frac{9}{sin ( x ) + cos ( x )})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{1}{sin ( x ) + cos ( x )})

Apply exponent rule:

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

= 9 \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)

= 9 (- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x)))

Simplify

9 (- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x))) : - \frac{9 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

= - \frac{9 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

(- 3, \frac{2}{5}), (4, \frac{3}{10})

h \to 0 lim (\frac{\frac{1}{1 + h} - 1}{h})

\int \frac{5}{6} x^{\frac{6}{5}} d x

f (x) = 12 x + x^{2} - x^{3}, g (x) = 0

What is the derivative of 9/(sin(x+cos(x))) ?
The derivative of 9/(sin(x+cos(x))) is -(9(cos(x)-sin(x)))/(1+sin(2x))
What is the first derivative of 9/(sin(x+cos(x))) ?
The first derivative of 9/(sin(x+cos(x))) is -(9(cos(x)-sin(x)))/(1+sin(2x))