What is the derivative of f(x)=arccsc(2x^4-5) ?

The derivative of f(x)=arccsc(2x^4-5) is -(4x^3)/(sqrt((2x^4-5)^2)\sqrt{x^8-5x^4+6)}

What is the first derivative of f(x)=arccsc(2x^4-5) ?

The first derivative of f(x)=arccsc(2x^4-5) is -(4x^3)/(sqrt((2x^4-5)^2)\sqrt{x^8-5x^4+6)}

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derivative of

f (x) = arccsc (2 x^{4} - 5)

- \frac{4 x ^{3}}{( 2 x ^{4} - 5 ) ^{2} x ^{8} - 5 x ^{4} + 6}

Solution steps

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

Apply the chain rule:

- \frac{1}{( 2 x ^{4} - 5 ) ^{2} ( 2 x ^{4} - 5 ) ^{2} - 1} \frac{d}{d x} (2 x^{4} - 5)

= - \frac{1}{( 2 x ^{4} - 5 ) ^{2} ( 2 x ^{4} - 5 ) ^{2} - 1} \frac{d}{d x} (2 x^{4} - 5)

\frac{d}{d x} (2 x^{4} - 5) = 8 x^{3}

= - \frac{1}{( 2 x ^{4} - 5 ) ^{2} ( 2 x ^{4} - 5 ) ^{2} - 1} \cdot 8 x^{3}

Simplify

- \frac{1}{( 2 x ^{4} - 5 ) ^{2} ( 2 x ^{4} - 5 ) ^{2} - 1} \cdot 8 x^{3} : - \frac{4 x ^{3}}{( 2 x ^{4} - 5 ) ^{2} x ^{8} - 5 x ^{4} + 6}

= - \frac{4 x ^{3}}{( 2 x ^{4} - 5 ) ^{2} x ^{8} - 5 x ^{4} + 6}

\int_{1}^{2} \frac{( x - 4 ) ^{2}}{x ^{5}} d x

(\frac{11}{8}, - \frac{1}{2}), (\frac{1}{4}, \frac{1}{4})

x \to 0 lim (\frac{sin ( 3 )}{2 x})

y = 3 x^{2}, (- 4, 48)

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

What is the derivative of f(x)=arccsc(2x^4-5) ?
The derivative of f(x)=arccsc(2x^4-5) is -(4x^3)/(sqrt((2x^4-5)^2)\sqrt{x^8-5x^4+6)}
What is the first derivative of f(x)=arccsc(2x^4-5) ?
The first derivative of f(x)=arccsc(2x^4-5) is -(4x^3)/(sqrt((2x^4-5)^2)\sqrt{x^8-5x^4+6)}