Math Snap

STEP 1

Assumptions
1. We are given two functions for which we need to find the derivative $\frac{dy}{dx}$.
2. The first function is $y = x \sin x$.
3. The second function is $y = \frac{x^2 - 1}{x^2 + 1}$.
4. We will use differentiation rules such as the product rule and the quotient rule.

STEP 2

For the first function $y = x \sin x$, we will use the product rule for differentiation. The product rule states that if $y = u \cdot v$, then $\frac{dy}{dx} = u'v + uv'$.

STEP 3

Identify $u$ and $v$ for the function $y = x \sin x$.
Let $u = x$ and $v = \sin x$.

STEP 4

Differentiate $u = x$ with respect to $x$.
$$u' = \frac{d}{dx}(x) = 1$$

STEP 5

Differentiate $v = \sin x$ with respect to $x$.
$$v' = \frac{d}{dx}(\sin x) = \cos x$$

STEP 6

Apply the product rule to find $\frac{dy}{dx}$ for $y = x \sin x$.
$$\frac{dy}{dx} = u'v + uv' = 1 \cdot \sin x + x \cdot \cos x$$

STEP 7

Simplify the expression for $\frac{dy}{dx}$.
$$\frac{dy}{dx} = \sin x + x \cos x$$

STEP 8

Now, move on to the second function $y = \frac{x^2 - 1}{x^2 + 1}$. We will use the quotient rule for differentiation. The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$.

STEP 9

Identify $u$ and $v$ for the function $y = \frac{x^2 - 1}{x^2 + 1}$.
Let $u = x^2 - 1$ and $v = x^2 + 1$.

STEP 10

Differentiate $u = x^2 - 1$ with respect to $x$.
$$u' = \frac{d}{dx}(x^2 - 1) = 2x$$

STEP 11

Differentiate $v = x^2 + 1$ with respect to $x$.
$$v' = \frac{d}{dx}(x^2 + 1) = 2x$$

STEP 12

Apply the quotient rule to find $\frac{dy}{dx}$ for $y = \frac{x^2 - 1}{x^2 + 1}$.
$$\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{(2x)(x^2 + 1) - (x^2 - 1)(2x)}{(x^2 + 1)^2}$$

STEP 13

Simplify the expression for $\frac{dy}{dx}$.
First, expand the terms in the numerator:
$$(2x)(x^2 + 1) = 2x^3 + 2x$$ $$(x^2 - 1)(2x) = 2x^3 - 2x$$

STEP 14

Subtract the second expression from the first in the numerator:
$$2x^3 + 2x - (2x^3 - 2x) = 2x^3 + 2x - 2x^3 + 2x = 4x$$

STEP 15

Write the simplified expression for $\frac{dy}{dx}$.
$$\frac{dy}{dx} = \frac{4x}{(x^2 + 1)^2}$$

The derivatives for the given functions are:
For $y = x \sin x$, $\frac{dy}{dx} = \sin x + x \cos x$.
For $y = \frac{x^2 - 1}{x^2 + 1}$, $\frac{dy}{dx} = \frac{4x}{(x^2 + 1)^2}$.