Could you please put the steps to it?
AS GIVEN the question MUST be read as :-
f ( x ) = (√5) x ² + 4
f ` ( x ) = (2√5) x
f " ( x ) = 2√5
Let a = 5x^2 + 4, so da/dx = 10x
Let b = sqrt(5x^2 + 4) = sqrt(a) = a^(1/2), so db/da = (1/2)a^(-1/2) = 1 / (2 * sqrt(5x^2 + 4))
By the Chain Rule:
db/dx = db/da * da/dx
db/dx = (1 / (2 * sqrt(5x^2 + 4))) * (10x)
db/dx = (5x) / sqrt(5x^2 + 4)
Let c = 5x, so dc/dx = 5
Since db/dx = c/b, use the Quotient Rule:
b'' = (c'b - cb') / (b^2)
b'' = ((5)(sqrt(5x^2 + 4)) - (5x)((5x) / sqrt(5x^2 + 4))) / ((sqrt(5x^2 + 4))^2)
b'' = (5*sqrt(5x^2 + 4) - ((25x^2) / sqrt(5x^2 + 4))) / (5x^2 + 4)
b'' = (5(5x^2 + 4) - 25x^2) / (sqrt(5x^2 + 4)^3)
b'' = (25x^2 + 20 - 25x^2) / sqrt(125x^6 + 300x^4 + 240x^2 + 64)
b'' = 20 / sqrt(125x^6 + 300x^4 + 240x^2 + 64)
f(x) = â5x^2+4
f '(x) = 2â5x
f "(x) = 2â5
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Verified answer
AS GIVEN the question MUST be read as :-
f ( x ) = (√5) x ² + 4
f ` ( x ) = (2√5) x
f " ( x ) = 2√5
Let a = 5x^2 + 4, so da/dx = 10x
Let b = sqrt(5x^2 + 4) = sqrt(a) = a^(1/2), so db/da = (1/2)a^(-1/2) = 1 / (2 * sqrt(5x^2 + 4))
By the Chain Rule:
db/dx = db/da * da/dx
db/dx = (1 / (2 * sqrt(5x^2 + 4))) * (10x)
db/dx = (5x) / sqrt(5x^2 + 4)
Let c = 5x, so dc/dx = 5
Since db/dx = c/b, use the Quotient Rule:
b'' = (c'b - cb') / (b^2)
b'' = ((5)(sqrt(5x^2 + 4)) - (5x)((5x) / sqrt(5x^2 + 4))) / ((sqrt(5x^2 + 4))^2)
b'' = (5*sqrt(5x^2 + 4) - ((25x^2) / sqrt(5x^2 + 4))) / (5x^2 + 4)
b'' = (5(5x^2 + 4) - 25x^2) / (sqrt(5x^2 + 4)^3)
b'' = (25x^2 + 20 - 25x^2) / sqrt(125x^6 + 300x^4 + 240x^2 + 64)
b'' = 20 / sqrt(125x^6 + 300x^4 + 240x^2 + 64)
f(x) = â5x^2+4
f '(x) = 2â5x
f "(x) = 2â5