What is the derivative of arctan((3x-8/(x^3))) ?

The derivative of arctan((3x-8/(x^3))) is (6x^2(-x+4))/((3x-8)^2+x^6)

What is the first derivative of arctan((3x-8/(x^3))) ?

The first derivative of arctan((3x-8/(x^3))) is (6x^2(-x+4))/((3x-8)^2+x^6)

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\frac{d}{d x} (arctan (\frac{3 x - 8}{x ^{3}}))

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

Solution steps

\frac{d}{d x} (arctan (\frac{3 x - 8}{x ^{3}}))

Apply the chain rule:

\frac{1}{( \frac{3 x - 8}{x ^{3}} ) ^{2} + 1} \frac{d}{d x} (\frac{3 x - 8}{x ^{3}})

= \frac{1}{( \frac{3 x - 8}{x ^{3}} ) ^{2} + 1} \frac{d}{d x} (\frac{3 x - 8}{x ^{3}})

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

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

Simplify

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

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

x \to \infty lim (x^{6})

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

x sin (- 2 x), at π

x \to 0 lim (\frac{x}{x ^{2} + 6 x})

\int t + 8 d t

What is the derivative of arctan((3x-8/(x^3))) ?
The derivative of arctan((3x-8/(x^3))) is (6x^2(-x+4))/((3x-8)^2+x^6)
What is the first derivative of arctan((3x-8/(x^3))) ?
The first derivative of arctan((3x-8/(x^3))) is (6x^2(-x+4))/((3x-8)^2+x^6)