Question

How to calculate: (f/g)'=?

derivations

Answer 1

`"see explanation"`

Explanation:

`"this is differentiated using the "color(blue)"quotient rule"`

`"given "y=(f(x))/(g(x))" then"`

`dy/dx=(f/g)^'=(g(x)f'(x)-f(x)g'(x))/(h(x))^2larrcolor(blue)"quotient rule"`

Answer 2

`(f/g)' = (f'g-fg')/g^2`

Explanation:

You can use the product rule and power rule to find:

`d/(dx) ((f(x))/(g(x))) = d/(dx) (f(x) * (g(x))^(-1))`

`color(white)(d/(dx) ((f(x))/(g(x)))) = (d/(dx) f(x)) * (g(x))^(-1) + f(x) * d/dx (g(x))^(-1)`

`color(white)(d/(dx) ((f(x))/(g(x)))) = (f'(x))/g(x) - f(x) * g'(x) * (g(x))^(-2)`

`color(white)(d/(dx) ((f(x))/(g(x)))) = (f'(x)g(x)-f(x)g'(x))/g(x)^2`

This is essentially the quotient rule:

`(f/g)' = (f'g-fg')/g^2`