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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{( 7 x ^{2} + x + 7 )}{( x ^{2} + 1 ) ^{2}} d x

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

\int_{0}^{4} e^{- x} d x

n = 1 \sum \infty (\frac{1}{4})^{n}

f (x) = x^{4} + 5 x

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