What is the derivative of f(x)=(7x)/(3+x^2) ?

The derivative of f(x)=(7x)/(3+x^2) is (7(3-x^2))/((3+x^2)^2)

What is the first derivative of f(x)=(7x)/(3+x^2) ?

The first derivative of f(x)=(7x)/(3+x^2) is (7(3-x^2))/((3+x^2)^2)

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derivative of

f (x) = \frac{7 x}{3 + x ^{2}}

\frac{7 ( 3 - x ^{2} )}{( 3 + x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{7 x}{3 + x ^{2}})

Take the constant out:

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

= 7 \frac{d}{d x} (\frac{x}{3 + x ^{2}})

Apply the Quotient Rule:

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

= 7 \cdot \frac{\frac{d x}{d x} ( 3 + x ^{2} ) - \frac{d}{d x} ( 3 + x ^{2} ) x}{( 3 + x ^{2} ) ^{2}}

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

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

= 7 \cdot \frac{1 \cdot ( 3 + x ^{2} ) - 2 xx}{( 3 + x ^{2} ) ^{2}}

Simplify

7 \cdot \frac{1 \cdot ( 3 + x ^{2} ) - 2 xx}{( 3 + x ^{2} ) ^{2}} : \frac{7 ( 3 - x ^{2} )}{( 3 + x ^{2} ) ^{2}}

= \frac{7 ( 3 - x ^{2} )}{( 3 + x ^{2} ) ^{2}}

y^{^{'}} + \frac{y}{x} - y = 0, y (1) = 0

\int (sec^{2} (x) - sin (x)) d x

x, x = 4

\int \frac{3 x + 1}{( x - 2 ) ( x + 3 )} d x

\int x - ln (x) d x

What is the derivative of f(x)=(7x)/(3+x^2) ?
The derivative of f(x)=(7x)/(3+x^2) is (7(3-x^2))/((3+x^2)^2)
What is the first derivative of f(x)=(7x)/(3+x^2) ?
The first derivative of f(x)=(7x)/(3+x^2) is (7(3-x^2))/((3+x^2)^2)