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

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

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

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

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derivative of

y = \frac{9 x ^{2} + 8}{x ^{2} + 1}

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

Solution steps

\frac{d}{d x} (\frac{9 x ^{2} + 8}{x ^{2} + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 9 x ^{2} + 8 ) ( x ^{2} + 1 ) - \frac{d}{d x} ( x ^{2} + 1 ) ( 9 x ^{2} + 8 )}{( x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (9 x^{2} + 8) = 18 x

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

= \frac{18 x ( x ^{2} + 1 ) - 2 x ( 9 x ^{2} + 8 )}{( x ^{2} + 1 ) ^{2}}

Expand

18 x (x^{2} + 1) - 2 x (9 x^{2} + 8) : 2 x

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

\int (\frac{6 x ^{5} - 3 x ^{2} + 4}{x ^{2}}) d x

\frac{d}{d x} (e^{s i n (x)})

- \frac{2 d y}{d x} + 6 x^{2} = 0

\int 6 x^{2} (2 x^{3} - 5)^{5} d x

\frac{d}{d x} (4 (1 - x)^{- 2})

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