AngouriMath
SolveAsTheDerivativeOfAProductOfPowersReadAtAPoint(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
SolveAsTheDerivativeOfAProductOfPowers(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) with the
constant read at a point where the brackets are proportional at the sampled points and
their quotient is not read free of x: Rubi's 1.1.3.3
(a - b x^(n/2))^p (a + b x^(n/2))^p (c + d x^n)^q at c = -a^2 d n (p + 1)/(b^2 (1 + n + n p)) and q = -(1 + 2 n + n p)/n is the derivative of
x (a - b x^(n/2))^(p + 1) (a + b x^(n/2))^(p + 1) (c + d x^n)^(q + 1) times
-b^2 (1 + n + n p)/(a^4 d n (p + 1)) , and was declined.
constant read at a point where the brackets are proportional at the sampled points and
their quotient is not read free of x: Rubi's 1.1.3.3
Remarks
Its bracket is a constant only once x x^(n - 1) is x^n and the symbolic
coefficients ofx^n and x^(2 n) cancel, which neither the reading off one
monomial nor the simplification of the quotient sees; atx = 1 the quotient is the
constant. Last in the chain and at the question asked only, so that what another rule
answers is answered as before; the answer is differentiated back at sampled points, as
the first's is.
https://github.com/asc-community/AngouriMath/issues/233
coefficients of
monomial nor the simplification of the quotient sees; at
constant. Last in the chain and at the question asked only, so that what another rule
answers is answered as before; the answer is differentiated back at sampled points, as
the first's is.
https://github.com/asc-community/AngouriMath/issues/233
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