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

The derivative of 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 sqrt((x^2-3x/(2x+1))) ?

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

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\frac{d}{d 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}}}

\frac{d}{d x} (sin (3 x^{4}))

\frac{d}{d x} (6 sin (x) - 8 e^{x})

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

f (x) = 2 x^{2} - 5 x, (- 1, 7)

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

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