Question

Writing to Learn Find the linearization of \(f(x)=\sqrt{x+1}+\sin x\) at \(x=0 .\) How is it related to the individual linearizations for \(\sqrt{x+1}\) and \(\sin x ?\)

Short answer

The linearization of the function \(f(x)=\sqrt{x+1}+\sin x\) at \(x=0\) is \(L(x) = 1 + \frac{3}{2}x\). The linearization of the sum of the functions is equivalent to the sum of the individual linearizations, which demonstrates the linearity of differentiation.

Step-by-step solution

Compute the Derivative

To linearize the function, first, take its derivative: \(f'(x) = \frac{1}{2\sqrt{x+1}} + \cos x\). Apply chain rule on \(\sqrt{x+1}\) to get \(\frac{1}{2\sqrt{x+1}}\) and the derivative of \(\sin x\) is \(\cos x\).

Evaluate the Derivative and Given Function at the point x = 0

To find the slope of the tangent line (which is essentially what linearization is), plug \(x=0\) into the derivative. This gives \(f'(0) = \frac{1}{2} + 1 = \frac{3}{2}\). Similarly, find \(f(0):\ \sqrt{0+1} + \sin 0 = 1 + 0 = 1\).

Form the linearization

The linearization of a function \(f(x)\) at a point \(a\) is given by \(f(a) + f'(a) * (x - a)\). Substituting \(a=0\), \(f(0) = 1\) and \(f'(0) = \frac{3}{2}\) to get the linearization, \(L(x) = 1 + \frac{3}{2} * x\).

Compare with individual linearization

If we were to find the individual linearizations of \(\sqrt{x+1}\) and \(\sin{x}\) at \(x=0\), we would obtain the functions \(L_1(x) = 1 + \frac{x}{2}\) and \(L_2(x) = x\), respectively. It is evident that the linearization of the sum of the functions (i.e. \(L(x)\)) is equivalent to the sum of the individual linearizations (i.e. \(L_1(x)\) and \(L_2(x)\)). This highlights the linearity of the derivative.

Key concepts

Derivative of a Function

The derivative of a function represents the rate at which the function's value changes as its input changes. More formally, if you have a function, say \(f(x)\), its derivative, labeled \(f'(x)\), tells us how \(f(x)\) reacts to small changes in \(x\). This concept is fundamental in calculus and has countless applications in mathematics, physics, engineering, and beyond.

When we compute the derivative of \(f(x) = \sqrt{x+1} + \sin x\), as in the given exercise, we're looking for the function that gives the slope of the tangent line to the curve at any point \(x\). To find this, we differentiate each part of the function separately, sometimes using rules like the chain rule for composite functions (explained in another section). In essence, \(f'(x)\) gives us a mathematical model for predicting how \(f\) will change near a given point.

Tangent Line Approximation

The tangent line approximation, also known as linearization, is a method of approximating the value of a function near a given point using the tangent line at that point. The idea is that if you zoom in closely enough on a curve at a point \(a\), the curve starts to look like a straight line. This straight line can be described by the equation \(L(x) = f(a) + f'(a) * (x - a)\), where \(f(a)\) is the function value at \(a\) and \(f'(a)\) is the slope of the tangent line at \(a\).

In our exercise, the tangent line approximation to \(f(x)\) at \(x=0\) yields \(L(x) = 1 + \frac{3}{2} * x\), indicating that very close to \(x=0\), the function \(f(x)\) can be closely approximated by this linear function. This method is indispensable for simplifying complex functions and studying their behavior near specific points.

Chain Rule

The chain rule is a formula for computing the derivative of the composition of two or more functions. When you have a function within another function, the chain rule enables you to find the derivative of the outer function with respect to the inner function and multiply it by the derivative of the inner function with respect to its variable.

For example, to find the derivative of \(\sqrt{x+1}\), we see this as an outer function \(\sqrt{u}\) with \(u = x+1\) being the inner function. Using the chain rule, we first differentiate \(\sqrt{u}\) with respect to \(u\) which gives \(\frac{1}{2\sqrt{u}}\), and then multiply by the derivative of \(u\) with respect to \(x\), which is 1. Thus, the derivative of \(\sqrt{x+1}\) is \(\frac{1}{2\sqrt{x+1}}\). This rule is essential when dealing with composite functions, making the process of finding derivatives systematic and more manageable.

Sum of Functions Derivative

The derivative of the sum of two functions is the sum of the derivatives of those functions. This is a direct consequence of the linearity of the derivative operation. To say it simply: if you want to differentiate \(f(x) + g(x)\), where \(f\) and \(g\) are functions of \(x\), you can just differentiate \(f\) and \(g\) separately and then add the results.

In our exercise, we differentiate \(\sqrt{x+1}\) and \(\sin x\) separately, obtaining their derivatives \(\frac{1}{2\sqrt{x+1}}\) and \(\cos x\) respectively. Accordingly, the derivative of the combined function \(f(x)\) is the sum of these two derivatives. This property allows us to decompose more complex functions into simpler parts, differentiate each part, and then combine the results to get the derivative of the whole.