Reciprocal Rule in Differentiation
How to solve the reciprocal rule in differentiation problems: formula, proof, example, and its solution.
[1/g(x)]' = -g'(x)/[g(x)]2
Write the denominator h at the front part. (= 1/h)
Solve 1/g(x + h) - 1/g(x).
Adding and subtracting rational expressions
Write (-) sign in front of 1/h.
And, to undo the (-) sign,
change g(x) - g(x + h)
to g(x + h) - g(x).
Combine 1/h and [g(x + h) - g(x)].
Then the remaining part is 1/[g(x + h)g(x)].
The limit of the gray part is g'(x).
And the limit of the latter part is 1/[g(x + 0)g(x)].
So [1/g(x)]' = -g'(x)/[g(x)]2.
y = 1/(x3 + 2x)
y' is equal to,
the derivative of x3 + 2x, (3x2 + 2⋅1x0)
over (x3 + 2x)2.
Derivatives of polynomials
So y' = -(3x2 + 2) / (x3 + 2x)2.