What is (\partial)/(\partial x)(ln(cos^2(x)+1)) ?

The answer to (\partial)/(\partial x)(ln(cos^2(x)+1)) is -(sin(2x))/(cos^2(x)+1)

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(\partial)/(\partial x)(ln(cos^2(x)+1))

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = \frac{x}{( 5 x - 4 ) ^{8}}, at x = 1

(- 2.2) (5.3)

y = x (x + 1)^{5}, (1, 32)

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

\frac{d y}{d x} + y^{9} x + 6 y = 0

What is (\partial)/(\partial x)(ln(cos^2(x)+1)) ?
The answer to (\partial)/(\partial x)(ln(cos^2(x)+1)) is -(sin(2x))/(cos^2(x)+1)