What is derivative of p=7q^3(5q-150) ?

The answer to derivative of p=7q^3(5q-150) is (dq)/(dp)= 1/(70q^2(2q-45))

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derivative of p=7q^3(5q-150)

derivative of

p = 7 q^{3} (5 q - 150)

\frac{d q}{d p} = \frac{1}{70 q ^{2} ( 2 q - 45 )}

Solution steps

p = 7 q^{3} (5 q - 150)

Treat

q

q (p)

Differentiate both sides:

1 = 7 (20 q^{3} \frac{d q}{d p} - 450 q^{2} \frac{d q}{d p})

Isolate

\frac{d q}{d p} : \frac{d q}{d p} = \frac{1}{70 q ^{2} ( 2 q - 45 )}

\frac{d q}{d p} = \frac{1}{70 q ^{2} ( 2 q - 45 )}

x^{2} + x y + y^{2} = 4

e^{x} y = 2 y

y^{3} - 3 x y^{2} = x^{3} + 3 x^{2} y

\frac{x ^{2}}{18} + \frac{y ^{2}}{8} = 1

x^{3} + y^{3} = 3 x y

What is derivative of p=7q^3(5q-150) ?
The answer to derivative of p=7q^3(5q-150) is (dq)/(dp)= 1/(70q^2(2q-45))