Question

Производная y=x^2/1+x

Answer 1

(2x(1+x)-x^2)/(1+x)^2
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Answer 2

To find the derivative of the function $y = \frac{x^2}{1 + x}$, we'll use the quotient rule, which states that if you have a function $f(x) = \frac{g(x)}{h(x)}$, then its derivative is given by:

$f'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{[h(x)]^2}$

Now, let's find the derivative of $y$ step by step:

Step 1: Identify $g(x)$ and $h(x)$: $g(x) = x^2$ $h(x) = 1 + x$

Step 2: Find $g'(x)$ and $h'(x)$: $g'(x) = \frac{d}{dx}(x^2) = 2x$ $h'(x) = \frac{d}{dx}(1 + x) = 1$

Step 3: Apply the quotient rule: $y' = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{[h(x)]^2} = \frac{(2x)(1 + x) - x^2 \cdot 1}{(1 + x)^2}$

Step 4: Simplify the expression: $y' = \frac{2x + 2x^2 - x^2}{(1 + x)^2} = \frac{x^2 + 2x}{(1 + x)^2}$

So, the derivative of $y = \frac{x^2}{1 + x}$ is $y' = \frac{x^2 + 2x}{(1 + x)^2}$.