What is the derivative of 1/(sin(x)cos(x+b+c)) ?

The derivative of 1/(sin(x)cos(x+b+c)) is -(cos(2x+b+c))/(cos^2(x+b+c)sin^2(x))

What is the first derivative of 1/(sin(x)cos(x+b+c)) ?

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

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derivative of

\frac{1}{sin ( x ) cos ( x + b + c )}

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

Solution steps

\frac{d}{d x} (\frac{1}{sin ( x ) cos ( x + b + c )})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (sin (x) cos (x + b + c)) = cos (2 x + b + c)

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

Simplify

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

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

\int_{0}^{1} arcsin (x) - sin (x) d x

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

f (x) = \frac{x ^{2}}{x + 9}

\int \frac{1}{cos ^{2} ( 7 x )} d x

(6 x^{2} + y - 1) d x - (14 y - x) d y = 0, y (1) = 0

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