What is (\partial)/(\partial x)(2/3 (1-x)^{3/2}) ?

The answer to (\partial)/(\partial x)(2/3 (1-x)^{3/2}) is -(1-x)^{1/2}

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(\partial)/(\partial x)(2/3 (1-x)^{3/2})

\frac{\partial}{\partial x} (\frac{2}{3} (1 - x)^{\frac{3}{2}})

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

Solution steps

\frac{\partial}{\partial x} (\frac{2}{3} (1 - x)^{\frac{3}{2}})

Take the constant out:

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

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

Apply the chain rule:

\frac{3 ( 1 - x ) ^{\frac{1}{2}}}{2} \frac{\partial}{\partial x} (1 - x)

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

\frac{\partial}{\partial x} (1 - x) = - 1

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

Simplify

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

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

n = 0 \sum \infty (\frac{1}{9})^{n}

\frac{d}{d x} (x + \frac{3}{x} + 5)

\frac{\partial}{\partial x} (tan (\frac{x}{2}))

y^{^{'''}} + 4 y^{^{''}} - 35 y^{^{'}} - 150 y = 0

What is (\partial)/(\partial x)(2/3 (1-x)^{3/2}) ?
The answer to (\partial)/(\partial x)(2/3 (1-x)^{3/2}) is -(1-x)^{1/2}