derivatives?

Question

1) y=2-x^2 divided by 3x^2+1


3) g(x)= 1 divided by 2x+3

Answer 1

it's been a few years since i've taken calculus...and i'm watching the spurs game now...i can tell you that you need to use the division rule w/ derivatives...

which you take the (top X deriv of bottom) - (bottom X deriv of top)

DIVIDED BY the bottom squared

and you simplify that out

Answer 2

If you mean the derivatives of the equations, then
f(x) = (2-x^2) / (3x^2+1)
quotient rule or laziness of ln
lny = ln ((2-x^2) / (3x^2+1)) = ln (2-x^2) - ln(3x^2+1)
y'/y = (-2x)/(2-x^2) - (6x)/(3x^2+1)
multiply by y which is the original function
[(2-x^2)[(-2x)/(2-x^2) - (6x)/(3x^2+1)]] / (3x^2+1)
then make sense of it all and simplify
......
g (x) = 1/ (2x+3)
g (x) = u^(-1)
u= (2x+3)
g' (x) = -u^(-2)
Replace u and multiply by the derivative of 2x +3
(-2) divided by (2x+3)^2

Answer 3

You would use the quotient rule.
f(x)=a(x)/b(x)
f'(x)=[a'(x)b(x) - b'(x)a(x)] / b(x)^2

Number 1:
[(2 - x^2)'(3x^2 + 1) - (3x^2 + 1)(2 - x^2)'] / (3x^2 + 1)^2

Number 3: You can also use power rule.
Power Rule:
1 / (2x + 3) is the same as (2x + 3)^-1
so g'(x) = (-1(2x+3)^-2)2

Answer 4

You have the formula for the quotient of functions, do you not?
f(x) = g(x)/h(x)
f'(x) = (g'(x)*h(x) - h'(x)*g(x))/(h(x))²
Just plug 'n play ☺
If you don't know how to take the derivative of things such as 2-x^2 or 3x^2+1 by now, you're already screwed. Plan on thaking this class over.

Doug

Answer 5

1) y = (2 - x^2) / (3x^2 + 1)

One way to solve this is to use the quotient rule. Recall that
[f(x)/g(x)]' = [ f'(x)g(x) - f(x)g'(x) ] / [ g(x)]^2

Therefore,

dy/dx = [ (-2x) (3x^2 + 1) - (2 - x^2)(6x) ] / (3x^2 + 1)^2
dy/dx = [ -6x^3 - 2x - 6x(2 - x^2) ] / (3x^2 + 1)^2
dy/dx = [ -6x^3 - 2x - 12x + 6x^3 ] / (3x^2 + 1)^2
dy/dx = [ -14x ] / (3x^2 + 1)^2

3) g(x) = 1/(2x + 3)

Same deal.

g'(x) = [ (0)(2x + 3) - (1)(2) ] / (2x + 3)^2
g'(x) =[ 0 - 2 ] / (2x + 3)^2
g'(x) = -2/(2x + 3)^2