What is the d/(dp)((q)(2p^2+q^2-1000)) ?

The d/(dp)((q)(2p^2+q^2-1000)) is 4qp

What is the first d/(dp)((q)(2p^2+q^2-1000)) ?

The first d/(dp)((q)(2p^2+q^2-1000)) is 4qp

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\frac{d}{d p} ((q) (2 p^{2} + q^{2} - 1000))

4 qp

Solution steps

\frac{d}{d p} ((q) (2 p^{2} + q^{2} - 1000))

Treat

q

as a constant

Take the constant out:

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

= q \frac{d}{d p} (2 p^{2} + q^{2} - 1000)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= q (\frac{d}{d p} (2 p^{2}) + \frac{d}{d p} (q^{2}) - \frac{d}{d p} (1000))

\frac{d}{d p} (2 p^{2}) = 4 p

\frac{d}{d p} (q^{2}) = 0

\frac{d}{d p} (1000) = 0

= q (4 p + 0 - 0)

Simplify

= 4 qp

\frac{3 ^{2} - 25}{3 - 6}

(338.76 + 84.69 x)^{2} + (- 206 - 64 x)^{s}

\frac{\partial}{\partial x} ((y) (sin (z) x - x))

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

a (e^{- k a^{2}} - 2 k e^{- k a^{2}} a^{2})

What is the d/(dp)((q)(2p^2+q^2-1000)) ?
The d/(dp)((q)(2p^2+q^2-1000)) is 4qp
What is the first d/(dp)((q)(2p^2+q^2-1000)) ?
The first d/(dp)((q)(2p^2+q^2-1000)) is 4qp