What is the derivative of y=((5x^2+9))/(x^2+1) ?

The derivative of y=((5x^2+9))/(x^2+1) is -(8x)/((x^2+1)^2)

What is the first derivative of y=((5x^2+9))/(x^2+1) ?

The first derivative of y=((5x^2+9))/(x^2+1) is -(8x)/((x^2+1)^2)

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derivative of

y = \frac{( 5 x ^{2} + 9 )}{x ^{2} + 1}

- \frac{8 x}{( x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5 x ^{2} + 9}{x ^{2} + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 5 x ^{2} + 9 ) ( x ^{2} + 1 ) - \frac{d}{d x} ( x ^{2} + 1 ) ( 5 x ^{2} + 9 )}{( x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (5 x^{2} + 9) = 10 x

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

= \frac{10 x ( x ^{2} + 1 ) - 2 x ( 5 x ^{2} + 9 )}{( x ^{2} + 1 ) ^{2}}

Simplify

\frac{10 x ( x ^{2} + 1 ) - 2 x ( 5 x ^{2} + 9 )}{( x ^{2} + 1 ) ^{2}} : - \frac{8 x}{( x ^{2} + 1 ) ^{2}}

= - \frac{8 x}{( x ^{2} + 1 ) ^{2}}

t \frac{d y}{d t} + t y = 1 - y

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

y = \frac{5}{2} - \frac{1}{x}, y = x

\int_{0}^{1} 6 (1 + x)^{4} d x

\int \frac{1}{x + 3 - 1} d x

What is the derivative of y=((5x^2+9))/(x^2+1) ?
The derivative of y=((5x^2+9))/(x^2+1) is -(8x)/((x^2+1)^2)
What is the first derivative of y=((5x^2+9))/(x^2+1) ?
The first derivative of y=((5x^2+9))/(x^2+1) is -(8x)/((x^2+1)^2)