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

The derivative of f(x)=((sqrt(x)-2))/((2x^2+3)) is (-6x^2+16xsqrt(x)+3)/(2sqrt(x)(2x^2+3)^2)

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

The first derivative of f(x)=((sqrt(x)-2))/((2x^2+3)) is (-6x^2+16xsqrt(x)+3)/(2sqrt(x)(2x^2+3)^2)

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derivative of

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

\frac{- 6 x ^{2} + 16 x x + 3}{2 x ( 2 x ^{2} + 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{\frac{1}{2 x} ( 2 x ^{2} + 3 ) - 4 x ( x - 2 )}{( 2 x ^{2} + 3 ) ^{2}} : \frac{- 6 x ^{2} + 16 x x + 3}{2 x ( 2 x ^{2} + 3 ) ^{2}}

= \frac{- 6 x ^{2} + 16 x x + 3}{2 x ( 2 x ^{2} + 3 ) ^{2}}

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

\frac{d}{d x} (ln (1 - 6 x))

- x^{2} + 3 x + 4, at x = 1

\frac{d y}{d x} = \frac{- e ^{x}}{8}

What is the derivative of f(x)=((sqrt(x)-2))/((2x^2+3)) ?
The derivative of f(x)=((sqrt(x)-2))/((2x^2+3)) is (-6x^2+16xsqrt(x)+3)/(2sqrt(x)(2x^2+3)^2)
What is the first derivative of f(x)=((sqrt(x)-2))/((2x^2+3)) ?
The first derivative of f(x)=((sqrt(x)-2))/((2x^2+3)) is (-6x^2+16xsqrt(x)+3)/(2sqrt(x)(2x^2+3)^2)