Question

Find the Y upon DX for the following functions why is equal to bracket X -1 bracket X -2 upon route x

Answer 1

Rewrite the equation as

`y = (x-1) (x-2) x^(-1/2)`

Then by the product rule, the derivative is

`(dy)/(dx) = (x-2) x^(-1/2) + (x-1) x^(-1/2) - \frac12 (x-1) (x-2) x^(-3/2)`

and we can factorize this as

`(dy)/(dx) = \frac12 x^(-3/2) \left(2 (x-2) x^(3/2-1/2) + 2 (x-1) x^(3/2-1/2) - (x-1) (x-2)\right)`

`(dy)/(dx) = \frac12 x^(-3/2) \left(2 (x-2) x + 2 (x-1) x - (x-1) (x-2)\right)`

`(dy)/(dx) = \frac12 x^(-3/2) (3x^2 - 3x - 2)`

`(dy)/(dx) = (3x^2 - 3x - 2)/(2x^(3/2))`

and optionally expanded once more (if only to match the provided "Ans") to

`(dy)/(dx) = \frac32 x^(2-3/2) - \frac32 x^(1-3/2) - x^(-3/2)`

`(dy)/(dx) = \frac32 x^(1/2) - \frac32 x^(-1/2) - x^(-3/2)`

`(dy)/(dx) = \frac32 \sqrt x - \frac3{2\sqrt x} - \frac1{√(x^3)}`