What is the derivative of f(x)=(4x)/(3x^2+1) ?

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

What is the first derivative of f(x)=(4x)/(3x^2+1) ?

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

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derivative of

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

\frac{d}{d x} (3 x^{2} + 1) = 6 x

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

Simplify

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

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

x \to 4 lim (\frac{7}{( x - 4 )})

\frac{\partial}{\partial x} (y ln (x y))

x \to 2 lim (\frac{6}{( x - 2 ) ^{3}})

\frac{d y}{d x}, 7 x + 4 y = 8

\int (5 e^{x} - \frac{2}{x}) d x

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