What is the derivative of arctan((a^2-x^3/(b^2))) ?

The derivative of arctan((a^2-x^3/(b^2))) is -(3b^2x^2)/((-x^3+a^2)^2+b^4)

What is the first derivative of arctan((a^2-x^3/(b^2))) ?

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

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\frac{d}{d x} (arctan (\frac{a ^{2} - x ^{3}}{b ^{2}}))

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

Solution steps

\frac{d}{d x} (arctan (\frac{a ^{2} - x ^{3}}{b ^{2}}))

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{a ^{2} - x ^{3}}{b ^{2}}) = - \frac{3 x ^{2}}{b ^{2}}

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

Simplify

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

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

\frac{d}{d x} (x \cdot (x^{5} + x^{4} + 3 x + 2))

\frac{\partial ^{2}}{\partial K ( b ) \partial b} (a K (b) (b)^{a} L^{b})

\frac{d y}{d x} - 6 x y = sin (x^{2})

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

\int 4 x - 3 x^{2} d x

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