What is the derivative of y=10csc(9x^4-9x+4) ?

The derivative of y=10csc(9x^4-9x+4) is -10cot(9x^4-9x+4)csc(9x^4-9x+4)(36x^3-9)

What is the first derivative of y=10csc(9x^4-9x+4) ?

The first derivative of y=10csc(9x^4-9x+4) is -10cot(9x^4-9x+4)csc(9x^4-9x+4)(36x^3-9)

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derivative of

y = 10 csc (9 x^{4} - 9 x + 4)

- 10 cot (9 x^{4} - 9 x + 4) csc (9 x^{4} - 9 x + 4) (36 x^{3} - 9)

Solution steps

\frac{d}{d x} (10 csc (9 x^{4} - 9 x + 4))

Take the constant out:

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

= 10 \frac{d}{d x} (csc (9 x^{4} - 9 x + 4))

Apply the chain rule:

- cot (9 x^{4} - 9 x + 4) csc (9 x^{4} - 9 x + 4) \frac{d}{d x} (9 x^{4} - 9 x + 4)

= - cot (9 x^{4} - 9 x + 4) csc (9 x^{4} - 9 x + 4) \frac{d}{d x} (9 x^{4} - 9 x + 4)

\frac{d}{d x} (9 x^{4} - 9 x + 4) = 36 x^{3} - 9

= 10 (- cot (9 x^{4} - 9 x + 4) csc (9 x^{4} - 9 x + 4) (36 x^{3} - 9))

Simplify

= - 10 cot (9 x^{4} - 9 x + 4) csc (9 x^{4} - 9 x + 4) (36 x^{3} - 9)

\frac{1}{s [ ( s + 2 ) ^{2} + 1 ]}

y^{^{'}} = \frac{x + 2 y - 1}{x + 2 y + 7}

f (x) = 8 x^{3} + 7 x

y^{^{'}} - y = 4

\int (x^{2} - 6 x) d x

What is the derivative of y=10csc(9x^4-9x+4) ?
The derivative of y=10csc(9x^4-9x+4) is -10cot(9x^4-9x+4)csc(9x^4-9x+4)(36x^3-9)
What is the first derivative of y=10csc(9x^4-9x+4) ?
The first derivative of y=10csc(9x^4-9x+4) is -10cot(9x^4-9x+4)csc(9x^4-9x+4)(36x^3-9)