Question

If y = 3-u\2+u and u = 4x\1-x^2;find dy\dx

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\frac{dy}{dx}$  =   $\frac{-5(1+x^2)}{(1-x^2+{2x}^2)(1-x^2)}$

Step-by-step explanation:

Given that y = $\frac{3-u}{2+u}$

Differentiating both sides with respect to u

=> $\frac{dy}{du}$ = $\frac{(2+u)(0-1)- (3-u)(0+1)}{(2+u)^2}$

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

        = $\frac{-2-u- 3+u}{(2+u)^2}$

        =  $\frac{-5}{(2+u)^2}$              

Also, it is given that u = $\frac{4x}{1-x^2}$

Differentiating both sides with respect to x

=> $\frac{du}{dx}$ = $\frac{(1-x^2)(4) - 4x (0-2x)}{(1-x^2)^2}$

        = $\frac{4(1-x^2) - 4x (-2x)}{(1-x^2)^2}$

        = $\frac{4-4x^2 +8x^2}{(1-x^2)^2}$

        = $\frac{4+4x^2}{(1-x^2)^2}$

But, we need to find $\frac{dy}{dx}$

Therefore,  $\frac{dy}{dx}$ = $\frac{dy}{du}$ *  $\frac{du}{dx}$

                       =  $\frac{-5}{(2+u)^2}$  *  $\frac{4+4x^2}{(1-x^2)^2}$

Substituting the value of u

$\frac{dy}{dx}$ =  $\frac{-5}{(2+\frac{4x}{1-x^2})^2}$  *  $\frac{4+4x^2}{(1-x^2)^2}$

    =   $\frac{-5(1-x^2)}{(2(1-x^2)+{4x})^2}$  *  $\frac{4+4x^2}{(1-x^2)^2}$

    =   $\frac{-5+5x^2}{(2-2x^2+{4x})^2}$  *  $\frac{4+4x^2}{(1-x^2)^2}$

    =   $\frac{-5(1-x^2)}{(2-2x^2+{4x})^2}$  *  $\frac{4+4x^2}{(1-x^2)^2}$

    =   $\frac{-5}{(2-2x^2+{4x})^2}$  * $\frac{4+4x^2}{(1-x^2)}$

    =   $\frac{-5}{2^2(1-x^2+{2x})^2}$  * $\frac{4(1+x^2)}{(1-x^2)}$

    =   $\frac{-5}{1-x^2+{2x}^2}$  * $\frac{1+x^2}{(1-x^2)}$

    =   $\frac{-5(1+x^2)}{(1-x^2+{2x}^2)(1-x^2)}$

Therefore, the value of  $\frac{dy}{dx}$  =   $\frac{-5(1+x^2)}{(1-x^2+{2x}^2)(1-x^2)}$