What is the derivative of 6/(1+cos(x)) ?

The derivative of 6/(1+cos(x)) is (6sin(x))/((1+cos(x))^2)

What is the first derivative of 6/(1+cos(x)) ?

The first derivative of 6/(1+cos(x)) is (6sin(x))/((1+cos(x))^2)

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\frac{d}{d x} (\frac{6}{1 + cos ( x )})

\frac{6 sin ( x )}{( 1 + cos ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6}{1 + cos ( x )})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{1}{1 + cos ( x )})

Apply exponent rule:

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

= 6 \frac{d}{d x} ((1 + cos (x))^{- 1})

Apply the chain rule:

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

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

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

= 6 (- \frac{1}{( 1 + cos ( x ) ) ^{2}} (- sin (x)))

Simplify

6 (- \frac{1}{( 1 + cos ( x ) ) ^{2}} (- sin (x))) : \frac{6 sin ( x )}{( 1 + cos ( x ) ) ^{2}}

= \frac{6 sin ( x )}{( 1 + cos ( x ) ) ^{2}}

x \to \infty lim (0)

y^{^{''}} + 8 y^{^{'}} + 16 y = 5 e^{- 4 t}

\int e^{- 19 x} d x

- 3 x - x^{2}, - 10

y^{^{'}} = 9 x + y, y (0) = 3

What is the derivative of 6/(1+cos(x)) ?
The derivative of 6/(1+cos(x)) is (6sin(x))/((1+cos(x))^2)
What is the first derivative of 6/(1+cos(x)) ?
The first derivative of 6/(1+cos(x)) is (6sin(x))/((1+cos(x))^2)