What is the derivative of f(x)=sqrt((3x+1)/(2x^2+1)) ?

The derivative of f(x)=sqrt((3x+1)/(2x^2+1)) is (-6x^2-4x+3)/(2sqrt(3x+1)(2x^2+1)^{3/2)}

What is the first derivative of f(x)=sqrt((3x+1)/(2x^2+1)) ?

The first derivative of f(x)=sqrt((3x+1)/(2x^2+1)) is (-6x^2-4x+3)/(2sqrt(3x+1)(2x^2+1)^{3/2)}

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derivative of

f (x) = \frac{3 x + 1}{2 x ^{2} + 1}

\frac{- 6 x ^{2} - 4 x + 3}{2 3 x + 1 ( 2 x ^{2} + 1 ) ^{\frac{3}{2}}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{3 x + 1}{2 x ^{2} + 1}) = \frac{- 6 x ^{2} - 4 x + 3}{( 2 x ^{2} + 1 ) ^{2}}

= \frac{1}{2 \frac{3 x + 1}{2 x ^{2} + 1}} \cdot \frac{- 6 x ^{2} - 4 x + 3}{( 2 x ^{2} + 1 ) ^{2}}

Simplify

\frac{1}{2 \frac{3 x + 1}{2 x ^{2} + 1}} \cdot \frac{- 6 x ^{2} - 4 x + 3}{( 2 x ^{2} + 1 ) ^{2}} : \frac{- 6 x ^{2} - 4 x + 3}{2 3 x + 1 ( 2 x ^{2} + 1 ) ^{\frac{3}{2}}}

= \frac{- 6 x ^{2} - 4 x + 3}{2 3 x + 1 ( 2 x ^{2} + 1 ) ^{\frac{3}{2}}}

\int \frac{1}{x ^{2} - 2 x + 50} d x

\int \frac{1}{x - 2 x ^{\frac{2}{3}} - 3 x ^{\frac{1}{3}}} d x

\int x 3 x^{2} - 5 d x

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

\int \frac{1}{3 x ^{2} + 6 x - 4} d x

What is the derivative of f(x)=sqrt((3x+1)/(2x^2+1)) ?
The derivative of f(x)=sqrt((3x+1)/(2x^2+1)) is (-6x^2-4x+3)/(2sqrt(3x+1)(2x^2+1)^{3/2)}
What is the first derivative of f(x)=sqrt((3x+1)/(2x^2+1)) ?
The first derivative of f(x)=sqrt((3x+1)/(2x^2+1)) is (-6x^2-4x+3)/(2sqrt(3x+1)(2x^2+1)^{3/2)}