What is the d/(dX)((8X^2+2X+8)/(sqrt(X))) ?

The d/(dX)((8X^2+2X+8)/(sqrt(X))) is (12X^2+X-4)/(Xsqrt(X))

What is the first d/(dX)((8X^2+2X+8)/(sqrt(X))) ?

The first d/(dX)((8X^2+2X+8)/(sqrt(X))) is (12X^2+X-4)/(Xsqrt(X))

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\frac{d}{d X} (\frac{8 X ^{2} + 2 X + 8}{X})

\frac{12 X ^{2} + X - 4}{X X}

Solution steps

\frac{d}{d X} (\frac{8 X ^{2} + 2 X + 8}{X})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d X} ( 8 X ^{2} + 2 X + 8 ) X - \frac{d}{d X} ( X ) ( 8 X ^{2} + 2 X + 8 )}{( X ) ^{2}}

\frac{d}{d X} (8 X^{2} + 2 X + 8) = 16 X + 2

\frac{d}{d X} (X) = \frac{1}{2 X}

= \frac{( 16 X + 2 ) X - \frac{1}{2 X} ( 8 X ^{2} + 2 X + 8 )}{( X ) ^{2}}

Simplify

\frac{( 16 X + 2 ) X - \frac{1}{2 X} ( 8 X ^{2} + 2 X + 8 )}{( X ) ^{2}} : \frac{12 X ^{2} + X - 4}{X X}

= \frac{12 X ^{2} + X - 4}{X X}

l a pl a ce t r an s f or m \frac{t - 5}{5}

\int sin^{3} (11 x) d x

\int_{- 3}^{3} (9 - x^{2}) d x

d er i v a t i v e \frac{π}{2}

x \to - 2 lim (\frac{x + 2}{x ^{2} - 4})

What is the d/(dX)((8X^2+2X+8)/(sqrt(X))) ?
The d/(dX)((8X^2+2X+8)/(sqrt(X))) is (12X^2+X-4)/(Xsqrt(X))
What is the first d/(dX)((8X^2+2X+8)/(sqrt(X))) ?
The first d/(dX)((8X^2+2X+8)/(sqrt(X))) is (12X^2+X-4)/(Xsqrt(X))