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

The derivative of 1/(sin(x+1)) is -cot(x+1)csc(x+1)

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

The first derivative of 1/(sin(x+1)) is -cot(x+1)csc(x+1)

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\frac{d}{d x} (\frac{1}{sin ( x + 1 )})

- cot (x + 1) csc (x + 1)

Solution steps

\frac{d}{d x} (\frac{1}{sin ( x + 1 )})

Simplify

\frac{1}{sin ( x + 1 )} : csc (x + 1)

= \frac{d}{d x} (csc (x + 1))

Apply the chain rule:

- cot (x + 1) csc (x + 1) \frac{d}{d x} (x + 1)

= - cot (x + 1) csc (x + 1) \frac{d}{d x} (x + 1)

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

= - cot (x + 1) csc (x + 1) \cdot 1

Simplify

= - cot (x + 1) csc (x + 1)

\int_{0}^{\frac{π}{2}} 3 sin^{2} (3 x) d x

\int \frac{3 x ^{2} + 5}{( x ^{2} + 4 ) ^{2}} d x

\int_{0}^{3} \frac{x}{36 - x ^{2}} d x

\int \frac{1 + x}{2 x} d x

y = x, y = 12 - 4 x, x = 0, x = 3

What is the derivative of 1/(sin(x+1)) ?
The derivative of 1/(sin(x+1)) is -cot(x+1)csc(x+1)
What is the first derivative of 1/(sin(x+1)) ?
The first derivative of 1/(sin(x+1)) is -cot(x+1)csc(x+1)