What is the derivative of (x^2/(sqrt(x^4+1))) ?

The derivative of (x^2/(sqrt(x^4+1))) is (2x)/((x^4+1)sqrt(x^4+1))

What is the first derivative of (x^2/(sqrt(x^4+1))) ?

The first derivative of (x^2/(sqrt(x^4+1))) is (2x)/((x^4+1)sqrt(x^4+1))

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (x^{2}) = 2 x

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

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

Simplify

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

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

\frac{d y}{d x} = 2 y - 1

x^{2} \frac{d y}{d x} + y^{2} = x y

\int cot^{2} (x) csc (x) d x

\frac{\partial}{\partial x} (- e^{2 y} cos (2 x) \cdot 2)

y = x^{4} - x^{2}, y^{2} = x^{2}

What is the derivative of (x^2/(sqrt(x^4+1))) ?
The derivative of (x^2/(sqrt(x^4+1))) is (2x)/((x^4+1)sqrt(x^4+1))
What is the first derivative of (x^2/(sqrt(x^4+1))) ?
The first derivative of (x^2/(sqrt(x^4+1))) is (2x)/((x^4+1)sqrt(x^4+1))