What is the derivative of y=8csc(2x^4-7x+3) ?

The derivative of y=8csc(2x^4-7x+3) is -8cot(2x^4-7x+3)csc(2x^4-7x+3)(8x^3-7)

What is the first derivative of y=8csc(2x^4-7x+3) ?

The first derivative of y=8csc(2x^4-7x+3) is -8cot(2x^4-7x+3)csc(2x^4-7x+3)(8x^3-7)

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derivative of

y = 8 csc (2 x^{4} - 7 x + 3)

- 8 cot (2 x^{4} - 7 x + 3) csc (2 x^{4} - 7 x + 3) (8 x^{3} - 7)

Solution steps

\frac{d}{d x} (8 csc (2 x^{4} - 7 x + 3))

Take the constant out:

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

= 8 \frac{d}{d x} (csc (2 x^{4} - 7 x + 3))

Apply the chain rule:

- cot (2 x^{4} - 7 x + 3) csc (2 x^{4} - 7 x + 3) \frac{d}{d x} (2 x^{4} - 7 x + 3)

= - cot (2 x^{4} - 7 x + 3) csc (2 x^{4} - 7 x + 3) \frac{d}{d x} (2 x^{4} - 7 x + 3)

\frac{d}{d x} (2 x^{4} - 7 x + 3) = 8 x^{3} - 7

= 8 (- cot (2 x^{4} - 7 x + 3) csc (2 x^{4} - 7 x + 3) (8 x^{3} - 7))

Simplify

= - 8 cot (2 x^{4} - 7 x + 3) csc (2 x^{4} - 7 x + 3) (8 x^{3} - 7)

\frac{d}{d x} (5 x^{5} f (x))

y^{^{'}} = e^{t + y}, y (0) = 0

\int 5 e^{3 x + 7} d x

\int_{0}^{5} (x^{2} - 5 x) d x

n = 0 \sum \infty (- 1)^{n} 2

What is the derivative of y=8csc(2x^4-7x+3) ?
The derivative of y=8csc(2x^4-7x+3) is -8cot(2x^4-7x+3)csc(2x^4-7x+3)(8x^3-7)
What is the first derivative of y=8csc(2x^4-7x+3) ?
The first derivative of y=8csc(2x^4-7x+3) is -8cot(2x^4-7x+3)csc(2x^4-7x+3)(8x^3-7)