What is differentiation of function. y= log(x+√(x2+1)) and define the (yn)0

To understand the differentiation of the function \( y = \log(x + \sqrt{x^2 + 1}) \), we first need to grasp what differentiation means in calculus. Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes as its input changes. In simpler terms, it tells us how steep the function is at any given point.

Breaking Down the Function

The function we are working with is \( y = \log(x + \sqrt{x^2 + 1}) \). This function involves a logarithm and a square root, both of which can complicate the differentiation process. To differentiate this function, we will use the chain rule and the properties of logarithms.

Step-by-Step Differentiation

Let's differentiate \( y \) with respect to \( x \). We can start by applying the chain rule, which states that if you have a composite function, the derivative is the derivative of the outer function multiplied by the derivative of the inner function.

• First, identify the inner function: \( u = x + \sqrt{x^2 + 1} \).
• Now, differentiate \( y \) with respect to \( u \): \( \frac{dy}{du} = \frac{1}{u} \) (since the derivative of \( \log(u) \) is \( \frac{1}{u} \)).
• Next, differentiate \( u \) with respect to \( x \):
  • For \( \sqrt{x^2 + 1} \), use the chain rule:
    • Let \( v = x^2 + 1 \), then \( \sqrt{v} \) has a derivative of \( \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx} = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \).
  • Thus, \( \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 + 1}} \).

Combining the Results

Now we can combine these results using the chain rule:

\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot \left(1 + \frac{x}{\sqrt{x^2 + 1}}\right) \)

Substituting back \( u = x + \sqrt{x^2 + 1} \), we get:

\( \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \left(1 + \frac{x}{\sqrt{x^2 + 1}}\right) \)

Understanding (yn)0

Now, regarding the notation \( (y_n)_0 \), it typically refers to a sequence or a specific value of the function \( y \) at a certain point, often at \( x = 0 \). To find \( y_0 \), we substitute \( x = 0 \) into our original function:

\( y(0) = \log(0 + \sqrt{0^2 + 1}) = \log(1) = 0 \)

So, \( (y_n)_0 = 0 \). This means that at \( x = 0 \), the value of the function \( y \) is 0.

Final Thoughts

In summary, we differentiated the function \( y = \log(x + \sqrt{x^2 + 1}) \) using the chain rule and found that the derivative is \( \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{x + \sqrt{x^2 + 1}} \). Additionally, we determined that \( (y_n)_0 = 0 \), indicating the value of the function at \( x = 0 \). Understanding these concepts is crucial for further studies in calculus and its applications.