let f(x)=6x^4-3x^5+5x^3/X^2 find f"(X)

Question

stuck on this problem, please help me?

Answer 1

(6x^4 - 3x^5 + 5x^3)/x^2 = 6x^2 - 3x^3 + 5x <--- factors out x^2 in the numerator, and it cancels out.


f'(x) = 12x - 9x^2 + 5

f''(x) = 12 - 18x

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Don't believe me... just watch ;-)

quotient rule:

f'(x)=

[ x^2 ( 24^3 - 15x^4 + 15x^2) - ( 6x^4 - 3x^5 + 5x^3) (2x) ]/x^4 =

[24x^5 - 15x^6 + 15x^4 - 12x^5 + 6x^6 - 10x^4 ]/ x^4

[12x^5 - 9x^6 +5x^4]/x^4

12x - 9x^2 + 5 <---- first derivatives agree via quotient rule.

Second derivative will also be the same as the denominator is gone.

Answer 2

F(x) = 6x⁴ -3x⁵ + 5x³/x²  is the original equation.  Simplify the 3rd term:

 

F(x) = 6x⁴ -3x⁵ + 5x¹  (note simple division of the last term.  Now calculate F'(x):

 

F'(x) = 24x³ -15x⁴ + 5....using the standard rules, now find F''(x):

 

F''(x) = 72x² -60x³  I'm assuming you were just confused by the last term.