Double Derivative of f(x) = (-5x^2+3x)/(2x^2-5)

[f22archrer]

Homework Statement

f(x) = (-5x^2+3x) / (2x^2-5)

Homework Equations

f 'x)=
(-6x^2+50x-15) / ( 2x^2-5)^2

The Attempt at a Solution

f ''x= ?

 

[berkeman]

Homework Statement

f(x) = (-5x^2+3x) / (2x^2-5)

Homework Equations

f 'x)=
(-6x^2+50x-15) / ( 2x^2-5)^2

The Attempt at a Solution

f ''x= ?

Can you please show all the steps you used to take the first derivative?

 

[f22archrer]

f'(x) = (2x^2 -5)((-10x+3) -(-5x^2+3x)4x) / 2x^2-5
= -20x^3 +6x^2+50x-15+20x^3-12x^2 / (2x^ - 5)^2
= -6x^2 +50x-15 / (2x^2 - 5)^2

 

[berkeman]

f'(x) = (2x^2 -5)((-10x+3) -(-5x^2+3x)4x) / 2x^2-5
= -20x^3 +6x^2+50x-15+20x^3-12x^2 / (2x^ - 5)^2
= -6x^2 +50x-15 / (2x^2 - 5)^2

Thanks, that makes it much easier to check. I think it's correct so far, now just apply the quotient rule one more time to get the second derivative...

 

[lurflurf]

Either use the quotient rule on your first derivative (which is right), or use the second derivative quotient rule, or use the product rule.

\left( \dfrac{u}{v} \right) ^{\prime \prime}=\dfrac{u^{\prime \prime} v^2-2u^\prime v v^\prime+2u (v^\prime)^2-u v^{\prime \prime}}{v^3}

 

[SammyS]

f'(x) = ((2x^2 -5)( -10x+3) -(-5x^2+3x)4x) / (2x^2-5) ²

= (-20x^3 +6x^2+50x-15+20x^3-12x^2 ) / (2x^ - 5)^2

= ( -6x^2 +50x-15 ) / (2x^2 - 5)^2

It helps to use sufficient number of parentheses. A little spacing can also help.