What is (\partial)/(\partial x)((75-x) 1/3 pix^2) ?

The answer to (\partial)/(\partial x)((75-x) 1/3 pix^2) is pix(-x+50)

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(\partial)/(\partial x)((75-x) 1/3 pix^2)

\frac{\partial}{\partial x} ((75 - x) \frac{1}{3} π x^{2})

π x (- x + 50)

Solution steps

\frac{\partial}{\partial x} ((75 - x) \frac{1}{3} π x^{2})

Take the constant out:

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

= \frac{1}{3} π \frac{\partial}{\partial x} ((75 - x) x^{2})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 75 - x, g = x^{2}

= \frac{1}{3} π (\frac{\partial}{\partial x} (75 - x) x^{2} + \frac{\partial}{\partial x} (x^{2}) (75 - x))

\frac{\partial}{\partial x} (75 - x) = - 1

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

= \frac{1}{3} π ((- 1) x^{2} + 2 x (75 - x))

Simplify

\frac{1}{3} π ((- 1) x^{2} + 2 x (75 - x)) : π x (- x + 50)

= π x (- x + 50)

\frac{\partial}{\partial x} (x ln (y) + \frac{1}{9} x^{3})

x - 3

x \to 9 - lim (\frac{7}{x - 9})

\int_{0}^{π} 7 cos^{6} (x) d x

y = x sin (\frac{4}{x})

What is (\partial)/(\partial x)((75-x) 1/3 pix^2) ?
The answer to (\partial)/(\partial x)((75-x) 1/3 pix^2) is pix(-x+50)