What is the derivative of f(x)=sqrt((4x+1)/(2x^2+1)) ?

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{4 x + 1}{2 x ^{2} + 1}) = \frac{- 8 x ^{2} - 4 x + 4}{( 2 x ^{2} + 1 ) ^{2}}

= \frac{1}{2 \frac{4 x + 1}{2 x ^{2} + 1}} \cdot \frac{- 8 x ^{2} - 4 x + 4}{( 2 x ^{2} + 1 ) ^{2}}

Simplify

\frac{1}{2 \frac{4 x + 1}{2 x ^{2} + 1}} \cdot \frac{- 8 x ^{2} - 4 x + 4}{( 2 x ^{2} + 1 ) ^{2}} : \frac{2 ( - 2 x ^{2} - x + 1 )}{4 x + 1 ( 2 x ^{2} + 1 ) ^{\frac{3}{2}}}

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

\int \frac{x - y}{x + y} d y

\int 9 x^{\frac{1}{2}} d x

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

x \to \infty lim (2 x - 3)

\frac{5}{x}, 4

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