What is the derivative of (6x/(x^2+36)) ?

The derivative of (6x/(x^2+36)) is (6(-x^2+36))/((x^2+36)^2)

What is the first derivative of (6x/(x^2+36)) ?

The first derivative of (6x/(x^2+36)) is (6(-x^2+36))/((x^2+36)^2)

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\frac{d}{d x} (\frac{6 x}{x ^{2} + 36})

\frac{6 ( - x ^{2} + 36 )}{( x ^{2} + 36 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{x ^{2} + 36})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{x ^{2} + 36})

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} + 36 ) - \frac{d}{d x} ( x ^{2} + 36 ) x}{( x ^{2} + 36 ) ^{2}}

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

\frac{d}{d x} (x^{2} + 36) = 2 x

= 6 \cdot \frac{1 \cdot ( x ^{2} + 36 ) - 2 xx}{( x ^{2} + 36 ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( x ^{2} + 36 ) - 2 xx}{( x ^{2} + 36 ) ^{2}} : \frac{6 ( - x ^{2} + 36 )}{( x ^{2} + 36 ) ^{2}}

= \frac{6 ( - x ^{2} + 36 )}{( x ^{2} + 36 ) ^{2}}

e^{- 3}

t \to 1 + lim (t^{\frac{t}{1 - t}})

\frac{d}{d x} (\frac{sin ( x )}{tan ( x )})

x \to \frac{π}{5} - lim (cot (5 x))

\frac{d}{d x} (\frac{- 2 sin ( 2 x )}{13})

What is the derivative of (6x/(x^2+36)) ?
The derivative of (6x/(x^2+36)) is (6(-x^2+36))/((x^2+36)^2)
What is the first derivative of (6x/(x^2+36)) ?
The first derivative of (6x/(x^2+36)) is (6(-x^2+36))/((x^2+36)^2)