Differentiate implicitly to find dp/dx
p^2 + p + 2x = 40
d/dx(p^2 + p + 2x) = d/dx(40)
2p(dp/dx) + (dp/dx) + 2*1 = 0
(2p + 1)(dy/dx) = -2
dp/dx = -2 / 2p + 1
Did I implicitly differentiate correctly? If not, how does this problem work? Please show the steps! Thanks :)
p^2 + p + 2x = 40
d/dx(p^2 + p + 2x) = d/dx(40)
2p(dp/dx) + (dp/dx) + 2*1 = 0
(2p + 1)(dy/dx) = -2
dp/dx = -2 / 2p + 1
Did I implicitly differentiate correctly? If not, how does this problem work? Please show the steps! Thanks :)
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I do it one term at a time:
D(p^2 + p + 2x) = D(40)
D(p^2) + D(p) + D(2x) = 0
2p dp + dp + 2 dx = 0
(2p+1) dp = -2 dx
dp/dx = -2 / (2p+1)
I find it easier that way
Doing it directly as a d/dx of "something" can get confusing when you have terms with both variables in it (this problem does not, but later, you will get them)
D(p^2 + p + 2x) = D(40)
D(p^2) + D(p) + D(2x) = 0
2p dp + dp + 2 dx = 0
(2p+1) dp = -2 dx
dp/dx = -2 / (2p+1)
I find it easier that way
Doing it directly as a d/dx of "something" can get confusing when you have terms with both variables in it (this problem does not, but later, you will get them)
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YES , but use brackets dp/dx = -2 /( 2p + 1)