What is the derivative of arccsc(x^4+1) ?

The derivative of arccsc(x^4+1) is -(4x^3)/(sqrt(x^4)\sqrt{(x^4+1)^2)sqrt(x^4+2)}

What is the first derivative of arccsc(x^4+1) ?

The first derivative of arccsc(x^4+1) is -(4x^3)/(sqrt(x^4)\sqrt{(x^4+1)^2)sqrt(x^4+2)}

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derivative of

arccsc (x^{4} + 1)

- \frac{4 x ^{3}}{x ^{4} ( x ^{4} + 1 ) ^{2} x ^{4} + 2}

Solution steps

\frac{d}{d x} (arccsc (x^{4} + 1))

Apply the chain rule:

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

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

\frac{d}{d x} (x^{4} + 1) = 4 x^{3}

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

Simplify

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

= - \frac{4 x ^{3}}{x ^{4} ( x ^{4} + 1 ) ^{2} x ^{4} + 2}

n \to \infty lim (2^{\frac{1}{n}})

y^{^{'}} + e^{x} y = e^{3 x} y^{- 2}

6 - x^{2}, x^{2} - 2

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

\int_{- 2}^{2} ln (x + 3) d x

What is the derivative of arccsc(x^4+1) ?
The derivative of arccsc(x^4+1) is -(4x^3)/(sqrt(x^4)\sqrt{(x^4+1)^2)sqrt(x^4+2)}
What is the first derivative of arccsc(x^4+1) ?
The first derivative of arccsc(x^4+1) is -(4x^3)/(sqrt(x^4)\sqrt{(x^4+1)^2)sqrt(x^4+2)}