Question

x^2+3x-9/x^2-9x+3
differentiate the following functions with respect to x
plsss help me ​

Answer 1

Step-by-step explanation:

let given function be y

y'= ( v*u' - u* v' ) /v^2

= (x^2-9x+3)*(2x+3)-(x^2+3x-9)*(2x-9) } / [ x^2-9x+3 ]^2

= [2x^3-18x^2+6x+3x^2-27x+9-(2x^3+6x^2-18x-9x^2-27x+81) ]/ [ x^2-9x+3 ]^2

={ - 12x^2 -14x-72 } / [ x^2-9x+3 ]^2

= -2 ( 6x^2+7x-36 ) / [ x^2-9x+3 ]^2

Answer 2

Answer:

$\frac{d}{dx}$  $\frac{x^{2} + 3x-9}{x^{2} -9x+3}$  =  $\frac{- 12(x^{2} -2x+6)} {(x^{2} -9x+3)^{2}}$

Step-by-step explanation:

We need to differentiate $\frac{x^{2} + 3x-9}{x^{2} -9x+3}$ with respect to x

We know that $\frac{d}{dx}$ ($\frac{N}{D}$) = $\frac{D *\frac{d}{dx} N -N *\frac{d}{dx} D }{D^{2}}$

where N = Function in the numerator

           D = Function in the denominator

Applying the formula in this question, we get,

$\frac{d}{dx}$  $\frac{x^{2} + 3x-9}{x^{2} -9x+3}$  =  $\frac{(x^{2} -9x+3)\frac{d}{dx} ( x^{2} + 3x-9) - ( x^{2} + 3x-9)\frac{d}{dx}(x^{2} -9x+3) }{(x^{2} -9x+3)^{2}}$

                   =  $\frac{(x^{2} -9x+3)(2x+3) - ( x^{2} + 3x-9)(2x-9) }{(x^{2} -9x+3)^{2}}$

                   =  $\frac{(2x^{3}+3x^{2} - 18x^{2} -27x+6x+9 - ( 2x^{3} - 9x^{2} + 6x^{2} -27x-18x+81)} {(x^{2} -9x+3)^{2}}$

                   =  $\frac{(2x^{3}- 15x^{2} -21x+9 - ( 2x^{3} - 3x^{2}-45x+81)} {(x^{2} -9x+3)^{2}}$

                   =  $\frac{(2x^{3}- 15x^{2} -21x+9 - 2x^{3} + 3x^{2}+45x-81)} {(x^{2} -9x+3)^{2}}$

                   =  $\frac{(- 12x^{2} +24x-72)} {(x^{2} -9x+3)^{2}}$

                   =  $\frac{- 12(x^{2} -2x+6)} {(x^{2} -9x+3)^{2}}$

Therefore, $\frac{d}{dx}$  $\frac{x^{2} + 3x-9}{x^{2} -9x+3}$  =  $\frac{- 12(x^{2} -2x+6)} {(x^{2} -9x+3)^{2}}$