What is the derivative of (pir^2(pir^2-1))/4 ?

The derivative of (pir^2(pir^2-1))/4 is (pir(2pir^2-1))/2

What is the first derivative of (pir^2(pir^2-1))/4 ?

The first derivative of (pir^2(pir^2-1))/4 is (pir(2pir^2-1))/2

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derivative of

\frac{π r ^{2} ( π r ^{2} - 1 )}{4}

\frac{π r ( 2 π r ^{2} - 1 )}{2}

Solution steps

\frac{d}{d r} (\frac{π r ^{2} ( π r ^{2} - 1 )}{4})

Take the constant out:

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

= \frac{π}{4} \frac{d}{d r} (r^{2} (π r^{2} - 1))

Apply the Product Rule:

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

f = r^{2}, g = π r^{2} - 1

= \frac{π}{4} (\frac{d}{d r} (r^{2}) (π r^{2} - 1) + \frac{d}{d r} (π r^{2} - 1) r^{2})

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

\frac{d}{d r} (π r^{2} - 1) = 2 π r

= \frac{π}{4} (2 r (π r^{2} - 1) + 2 π r r^{2})

Simplify

\frac{π}{4} (2 r (π r^{2} - 1) + 2 π r r^{2}) : \frac{π r ( 2 π r ^{2} - 1 )}{2}

= \frac{π r ( 2 π r ^{2} - 1 )}{2}

\frac{\partial}{\partial x} (x y e^{- x^{2} - y^{2}})

f (x) = (6 x - 8)^{\frac{1}{2}}, at x = 4

\int_{0}^{π} x^{2} cos (n x) d x

\int \frac{1}{x ^{2} x ^{2} - 7} d x

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

What is the derivative of (pir^2(pir^2-1))/4 ?
The derivative of (pir^2(pir^2-1))/4 is (pir(2pir^2-1))/2
What is the first derivative of (pir^2(pir^2-1))/4 ?
The first derivative of (pir^2(pir^2-1))/4 is (pir(2pir^2-1))/2