Find the Derivative (dy/dx) of the following problem:
*will give medal*

Question

livya15 · Answer

$$f(x)=\frac{ x ^{3}+3x+2 }{ x ^{2}-1 }$$

livya15 · Answer

@mathstudent55 @amistre64

livya15 · Answer

@kohai

amistre64 · Answer

seems like a quotient rule to me

amistre64 · Answer

or run a product rule and factor the square of the bottom afterwards

livya15 · Answer

is this the answer: $$x ^{4}-6x ^{2}-4x-3$$

livya15 · Answer

@amistre64

amistre64 · Answer

$$f(x)=\frac{ x ^{3}+3x+2 }{ x ^{2}-1 }$$
$$f(x)=( x ^{3}+3x+2)~ (x ^{2}-1)^{-1}$$
$$f'(x)=( x ^{3}+3x+2)'~ (x ^{2}-1)^{-1}+( x ^{3}+3x+2)~ (x ^{2}-1)'^{-1}$$
$$f'(x)=( 3x ^{2}+3)~ (x ^{2}-1)^{-1}-2x( x ^{3}+3x+2)~ (x ^{2}-1)^{-2}$$
$$f'(x)=[( 3x ^{2}+3)~ (x ^{2}-1)-2x( x ^{3}+3x+2)]~ (x ^{2}-1)^{-2}$$
$$f'(x)=\frac{( 3x ^{2}+3)~ (x ^{2}-1)-2x( x ^{3}+3x+2)}{ (x ^{2}-1)^{2}}$$

amistre64 · Answer

3x^4 -3x^2 +3x^2 -3 -2x^4 -6x^2 -4

x^4  -6x^2 -7  seems to simplify the top

livya15 · Answer

thanks

amistre64 · Answer

good luck :)