What is the derivative of y=2csc(4x^4-2x+5) ?

The derivative of y=2csc(4x^4-2x+5) is -2cot(4x^4-2x+5)csc(4x^4-2x+5)(16x^3-2)

What is the first derivative of y=2csc(4x^4-2x+5) ?

The first derivative of y=2csc(4x^4-2x+5) is -2cot(4x^4-2x+5)csc(4x^4-2x+5)(16x^3-2)

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derivative of

y = 2 csc (4 x^{4} - 2 x + 5)

- 2 cot (4 x^{4} - 2 x + 5) csc (4 x^{4} - 2 x + 5) (16 x^{3} - 2)

Solution steps

\frac{d}{d x} (2 csc (4 x^{4} - 2 x + 5))

Take the constant out:

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

= 2 \frac{d}{d x} (csc (4 x^{4} - 2 x + 5))

Apply the chain rule:

- cot (4 x^{4} - 2 x + 5) csc (4 x^{4} - 2 x + 5) \frac{d}{d x} (4 x^{4} - 2 x + 5)

= - cot (4 x^{4} - 2 x + 5) csc (4 x^{4} - 2 x + 5) \frac{d}{d x} (4 x^{4} - 2 x + 5)

\frac{d}{d x} (4 x^{4} - 2 x + 5) = 16 x^{3} - 2

= 2 (- cot (4 x^{4} - 2 x + 5) csc (4 x^{4} - 2 x + 5) (16 x^{3} - 2))

Simplify

= - 2 cot (4 x^{4} - 2 x + 5) csc (4 x^{4} - 2 x + 5) (16 x^{3} - 2)

x^{^{''}} + 9 x = 0

n = 1 \sum \infty \frac{1}{1 + n ^{2}}

\frac{\partial}{\partial x} (2 x^{3} + y^{2})

\frac{d}{d z} (\frac{1}{1 - z ^{- 1}})

\frac{d}{d y} (4 \cdot x^{2} \cdot y^{3} - 4)

What is the derivative of y=2csc(4x^4-2x+5) ?
The derivative of y=2csc(4x^4-2x+5) is -2cot(4x^4-2x+5)csc(4x^4-2x+5)(16x^3-2)
What is the first derivative of y=2csc(4x^4-2x+5) ?
The first derivative of y=2csc(4x^4-2x+5) is -2cot(4x^4-2x+5)csc(4x^4-2x+5)(16x^3-2)