Problem 36

Question

Assume that $f(x)$ and $g(x)$ are differentiable. Find $\frac{d}{d x} \sqrt{f(x)+g(x)}$.

Step-by-Step Solution

Verified

Answer

$ \frac{1}{2\sqrt{f(x) + g(x)}} \cdot (f'(x) + g'(x)) $.

1Step 1: Use the Chain Rule

To find the derivative of $ \sqrt{f(x) + g(x)} $, recognize that it is a composition of functions. Specifically, it's the square root function applied to $ f(x) + g(x) $. Let $ u = f(x) + g(x) $. This makes the function $ \sqrt{u} $. The derivative of $ \sqrt{u} $ with respect to $ u $ is $ \frac{1}{2\sqrt{u}} $. We will apply the chain rule: $ \frac{d}{d x} \sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} $.

2Step 2: Differentiate f(x) + g(x)

Differentiate $ u = f(x) + g(x) $ with respect to $ x $. Since both $ f(x) $ and $ g(x) $ are differentiable, we use the sum rule, resulting in $ \frac{du}{dx} = f'(x) + g'(x) $.

3Step 3: Combine Results

Substitute the expression for $ \frac{du}{dx} $ into the chain rule result. This gives us: \[ \frac{d}{dx} \sqrt{f(x) + g(x)} = \frac{1}{2\sqrt{f(x) + g(x)}} \cdot (f'(x) + g'(x)) \].

Key Concepts

Chain RuleSum RuleDifferentiable Functions

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate a composition of functions. Imagine you have function g(x) nested inside another function f(x). When you differentiate the outer function, you must also apply the derivative of the inner function.
For our problem with the expression $\sqrt{f(x) + g(x)}$, we recognize it as a composition because of the square root applied to $f(x) + g(x)$.

First, we simplify the inner function by letting $u = f(x) + g(x)$. Now, the expression becomes $\sqrt{u}$.
The derivative of $\sqrt{u}$ in terms of $u$ is $\frac{1}{2\sqrt{u}}$.
According to the chain rule, the derivative concerning $x$ is $\frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$.

This multiplication of derivatives is crucial because it accounts for the rate of change not just of $\sqrt{u}$, but of the whole composition of functions.

Sum Rule

The sum rule simplifies the process of finding the derivative of a sum of functions. If you have two differentiable functions, their sum is also differentiable, and the derivative is simply the sum of their individual derivatives.
In our scenario, we want the derivative of $f(x) + g(x)$.

The sum rule tells us that $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$.

This rule comes in especially handy for simplifying complex derivatives into manageable pieces.

Differentiable Functions

A function is said to be differentiable if it has a derivative at each point in its domain. This means that its graph will have a tangent at each point and won't have any sharp corners or cusps.
In calculus, when a problem states that $f(x)$ and $g(x)$ are differentiable, it confirms that:

Both functions have a derivative at every point we're considering.
We can use rules like the chain rule or sum rule without any issue.
The functions are smooth enough for calculus procedures to apply.

Thus, recognizing that $f(x)$ and $g(x)$ are differentiable ensures that operations like integration and differentiation are seamless.

Problem 36

Other exercises in this chapter

Problem 36

A measurement error in $x$ affects the accuracy of the value $f(x) .$ In each case, determine an interval of the form $$[f(x)-\Delta f, f(x)+\Delta f]$$ tha

View solution

Problem 36

The following limit represents the derivative of a function $f$ at the point $(a, f(a))$ : $$ \lim _{h \rightarrow 0} \frac{4(a+h)^{3}-4 a^{3}}{h} $$

View solution

Problem 36

In Problems $1-58$, find the derivative with respect to the independent variable. $$ f(x)=\sin x \cos x $$

View solution

Problem 36

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=3^{x-1} $$

View solution