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

The derivative of f(x)=(x^2)/(x^2+1) is (2x)/((x^2+1)^2)

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

The first derivative of f(x)=(x^2)/(x^2+1) is (2x)/((x^2+1)^2)

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

x \to 0 + lim ((1 + 2 x)^{\frac{3}{x}})

(2 x - 7)^{5}

y^{^{''}} + 4 y = 2 x e^{3 x}, y (0) = 0, y^{^{'}} (0) = 0

y = 8 x^{2} - 6

f (x) = \frac{2}{4} x^{5} - \frac{1}{3} x^{3} + \frac{3}{6} x^{2} + \frac{1}{2} x

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