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

The derivative of f(x)=((5x-2))/((x^2+1)) is (-5x^2+4x+5)/((x^2+1)^2)

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

The first derivative of f(x)=((5x-2))/((x^2+1)) is (-5x^2+4x+5)/((x^2+1)^2)

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derivative of

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

\frac{- 5 x ^{2} + 4 x + 5}{( x ^{2} + 1 ) ^{2}}

Solution steps

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

\frac{d}{d x} (5 x - 2) = 5

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

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

Expand

5 (x^{2} + 1) - 2 x (5 x - 2) : - 5 x^{2} + 4 x + 5

= \frac{- 5 x ^{2} + 4 x + 5}{( x ^{2} + 1 ) ^{2}}

\int_{0}^{\frac{π}{8}} sec^{2} (2 x) d x

\int_{0}^{\frac{π}{6}} 1

x \to 0 - lim (x^{s i n (x)})

y = (1 + x)^{3}

\frac{d}{d t} (sec^{2} (t))

What is the derivative of f(x)=((5x-2))/((x^2+1)) ?
The derivative of f(x)=((5x-2))/((x^2+1)) is (-5x^2+4x+5)/((x^2+1)^2)
What is the first derivative of f(x)=((5x-2))/((x^2+1)) ?
The first derivative of f(x)=((5x-2))/((x^2+1)) is (-5x^2+4x+5)/((x^2+1)^2)