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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{\partial}{\partial y} (x cos (x + 2 y))

f (x) = \frac{1}{x + 1}, at x = 4

\int \frac{1}{( x ^{2} - 4 )} d x

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

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