What is the derivative of 2/3 ln(1+3cos^2(x)) ?

The derivative of 2/3 ln(1+3cos^2(x)) is -(2sin(2x))/(1+3cos^2(x))

What is the first derivative of 2/3 ln(1+3cos^2(x)) ?

The first derivative of 2/3 ln(1+3cos^2(x)) is -(2sin(2x))/(1+3cos^2(x))

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\frac{d}{d x} (\frac{2}{3} ln (1 + 3 cos^{2} (x)))

- \frac{2 sin ( 2 x )}{1 + 3 cos ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{2}{3} ln (1 + 3 cos^{2} (x)))

Take the constant out:

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

= \frac{2}{3} \frac{d}{d x} (ln (1 + 3 cos^{2} (x)))

Apply the chain rule:

\frac{1}{1 + 3 cos ^{2} ( x )} \frac{d}{d x} (1 + 3 cos^{2} (x))

= \frac{1}{1 + 3 cos ^{2} ( x )} \frac{d}{d x} (1 + 3 cos^{2} (x))

\frac{d}{d x} (1 + 3 cos^{2} (x)) = - 3 sin (2 x)

= \frac{2}{3} \cdot \frac{1}{1 + 3 cos ^{2} ( x )} (- 3 sin (2 x))

Simplify

\frac{2}{3} \cdot \frac{1}{1 + 3 cos ^{2} ( x )} (- 3 sin (2 x)) : - \frac{2 sin ( 2 x )}{1 + 3 cos ^{2} ( x )}

= - \frac{2 sin ( 2 x )}{1 + 3 cos ^{2} ( x )}

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

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

f (x) = x^{2} - 2, at x = 1

\frac{d}{d x} (6 x^{3} e^{x})

\int_{0}^{\frac{π}{4}} 2 sec^{2} (t) d t

What is the derivative of 2/3 ln(1+3cos^2(x)) ?
The derivative of 2/3 ln(1+3cos^2(x)) is -(2sin(2x))/(1+3cos^2(x))
What is the first derivative of 2/3 ln(1+3cos^2(x)) ?
The first derivative of 2/3 ln(1+3cos^2(x)) is -(2sin(2x))/(1+3cos^2(x))