What is the derivative of y=(6x)/(x^2+1) ?

The derivative of y=(6x)/(x^2+1) is (6(-x^2+1))/((x^2+1)^2)

What is the first derivative of y=(6x)/(x^2+1) ?

The first derivative of y=(6x)/(x^2+1) is (6(-x^2+1))/((x^2+1)^2)

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derivative of

y = \frac{6 x}{x ^{2} + 1}

\frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

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

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

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

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

Simplify

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

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

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

y = x^{2} + 3, (- 1, 4)

\int_{- 4}^{4} ∣ x ∣ d x

x \to - 3 lim (\frac{4 x - 5}{3 x + 9})

x \to 1 lim ([3 x + 1])

What is the derivative of y=(6x)/(x^2+1) ?
The derivative of y=(6x)/(x^2+1) is (6(-x^2+1))/((x^2+1)^2)
What is the first derivative of y=(6x)/(x^2+1) ?
The first derivative of y=(6x)/(x^2+1) is (6(-x^2+1))/((x^2+1)^2)