What is the derivative of (8x+3/(0.2x^2+1)) ?

The derivative of (8x+3/(0.2x^2+1)) is (-1.6x^2-1.2x+8)/((0.2x^2+1)^2)

What is the first derivative of (8x+3/(0.2x^2+1)) ?

The first derivative of (8x+3/(0.2x^2+1)) is (-1.6x^2-1.2x+8)/((0.2x^2+1)^2)

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\frac{d}{d x} (\frac{8 x + 3}{0.2 x ^{2} + 1})

\frac{- 1.6 x ^{2} - 1.2 x + 8}{( 0.2 x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{8 x + 3}{0.2 x ^{2} + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 8 x + 3 ) ( 0.2 x ^{2} + 1 ) - \frac{d}{d x} ( 0.2 x ^{2} + 1 ) ( 8 x + 3 )}{( 0.2 x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (8 x + 3) = 8

\frac{d}{d x} (0.2 x^{2} + 1) = 0.4 x

= \frac{8 ( 0.2 x ^{2} + 1 ) - 0.4 x ( 8 x + 3 )}{( 0.2 x ^{2} + 1 ) ^{2}}

Expand

8 (0.2 x^{2} + 1) - 0.4 x (8 x + 3) : - 1.6 x^{2} - 1.2 x + 8

= \frac{- 1.6 x ^{2} - 1.2 x + 8}{( 0.2 x ^{2} + 1 ) ^{2}}

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

\int \frac{x ^{6}}{( 1 + x ^{2} ) ^{2}} d x

(1 - y^{2}) d x - x y d y = 0

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

y^{^{'}} = \frac{t}{y e ^{- y^{2}} + y e ^{- y^{2}} t ^{2}}

What is the derivative of (8x+3/(0.2x^2+1)) ?
The derivative of (8x+3/(0.2x^2+1)) is (-1.6x^2-1.2x+8)/((0.2x^2+1)^2)
What is the first derivative of (8x+3/(0.2x^2+1)) ?
The first derivative of (8x+3/(0.2x^2+1)) is (-1.6x^2-1.2x+8)/((0.2x^2+1)^2)