What is the derivative of (6x/(x^2+3)) ?

The derivative of (6x/(x^2+3)) is (6(-x^2+3))/((x^2+3)^2)

What is the first derivative of (6x/(x^2+3)) ?

The first derivative of (6x/(x^2+3)) is (6(-x^2+3))/((x^2+3)^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{x - 7}{( x + 5 ) ^{10}})

x^{2} - 12 x + k

x \to 4 lim (x 25 - x^{2})

y^{^{''}} + 4 y^{^{'}} + 12 y = 0

\int_{0}^{4} (13 - t) t d t

What is the derivative of (6x/(x^2+3)) ?
The derivative of (6x/(x^2+3)) is (6(-x^2+3))/((x^2+3)^2)
What is the first derivative of (6x/(x^2+3)) ?
The first derivative of (6x/(x^2+3)) is (6(-x^2+3))/((x^2+3)^2)