Question

y= x /1+x² then dy /dx = a. (1 - x²)/(1 + x²) b. ((1 - x²)²/(1 + x²) C. (1 - x²)/(1 + x²)² d. None of these​

Answer 1

Answer:

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

Step-by-step explanation:

Given y = $\frac{x}{1+x^2}$

Using Quotient rule

$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)\frac{d}{dx}(f(x))-f(x)\frac{d}{dx}(g(x)) }{g(x)^2}$

Here  $y = \frac{x}{1+x^2}$

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

$\frac{dy}{dx} = \frac{(1+x^2)(1)-(x)(0+2x)}{(1+x^2)^2}$

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

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

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

Hence the derivative of  $\frac{x}{(1+x)^2}$ is  $\frac{(1-x)^2}{(1+x^2)^2}$

Option c

Formula used

$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)\frac{d}{dx}(f(x))-f(x)\frac{d}{dx}(g(x)) }{g(x)^2}$

                 OR

$\frac{d}{dx}(\frac{f(x)}{g(x)}) = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2}$