What is the derivative of (x^2/(6+sqrt(x))) ?

The derivative of (x^2/(6+sqrt(x))) is (4x(6+sqrt(x))-x^{3/2})/(2(6+sqrt(x))^2)

What is the first derivative of (x^2/(6+sqrt(x))) ?

The first derivative of (x^2/(6+sqrt(x))) is (4x(6+sqrt(x))-x^{3/2})/(2(6+sqrt(x))^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{2}) = 2 x

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

= \frac{2 x ( 6 + x ) - \frac{1}{2 x} x ^{2}}{( 6 + x ) ^{2}}

Simplify

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

= \frac{4 x ( 6 + x ) - x ^{\frac{3}{2}}}{2 ( 6 + x ) ^{2}}

f (x) = \frac{1}{x + 2}, at x = 3

\frac{\partial}{\partial x} (\frac{1}{1 - x ^{4}})

(3 + x) d x = (e^{t} - e^{- t}) d t

\int (x^{1.1} + 9 x^{3.5}) d x

\int 2 x^{2} cos (x) d x

What is the derivative of (x^2/(6+sqrt(x))) ?
The derivative of (x^2/(6+sqrt(x))) is (4x(6+sqrt(x))-x^{3/2})/(2(6+sqrt(x))^2)
What is the first derivative of (x^2/(6+sqrt(x))) ?
The first derivative of (x^2/(6+sqrt(x))) is (4x(6+sqrt(x))-x^{3/2})/(2(6+sqrt(x))^2)