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

The derivative of (x^4+1/(x^2+1)) is (2x^5+4x^3-2x)/((x^2+1)^2)

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

The first derivative of (x^4+1/(x^2+1)) is (2x^5+4x^3-2x)/((x^2+1)^2)

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

\frac{2 x ^{5} + 4 x ^{3} - 2 x}{( x ^{2} + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

4 x^{3} (x^{2} + 1) - 2 x (x^{4} + 1) : 2 x^{5} + 4 x^{3} - 2 x

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

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

\frac{d}{d x} (cos (e^{t a n (3 x)}))

\int 6 sin (6 x) d x

\frac{\partial}{\partial x} (e^{6})

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

What is the derivative of (x^4+1/(x^2+1)) ?
The derivative of (x^4+1/(x^2+1)) is (2x^5+4x^3-2x)/((x^2+1)^2)
What is the first derivative of (x^4+1/(x^2+1)) ?
The first derivative of (x^4+1/(x^2+1)) is (2x^5+4x^3-2x)/((x^2+1)^2)