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Chapter 3: Problem 9

Evaluate the derivative of \(y=\sqrt{4 x+1}\) using \(d / d x\left(f(g(x))=f^{\prime}(g(x)) \cdot g^{\prime}(x)\right.\).

Short Answer

Expert verified
The derivative of the function \(y=\sqrt{4x+1}\) is \(\frac{dy}{dx}=\frac{2}{\sqrt{4x+1}}\).

Step by step solution

01

Identify the inner and outer functions

Let's identify the inner function, \(g(x)\), and the outer function, \(f(x)\). In this case, the given function is \(y=\sqrt{4x+1}= (\sqrt{\vphantom{w}u})\) with \(u = 4x+1\). So, the inner function, \(g(x)=4x+1\), and the outer function, \(f(u)=\sqrt{\vphantom{w}u}\). Notice that \(f(g(x)) = \sqrt{4x+1}\).
02

Compute the derivatives of the inner and outer functions

Now let's find the derivatives of the inner and outer functions, denoted as \(f'(u)\) and \(g'(x)\), respectively. For the outer function \(f(u) = \sqrt{u}\), $$ f^{\prime}(u) = \frac{d}{du}(\sqrt{u}) = \frac{1}{2\sqrt{u}}. $$ For the inner function \(g(x) = 4x + 1\), $$ g^{\prime}(x) = \frac{d}{dx}(4x + 1) = 4. $$
03

Use the chain rule formula to find the derivative

Now we will apply the chain rule formula: $$ \frac{d}{dx}(f(g(x)))=f^{\prime}(g(x)) \cdot g^{\prime}(x). $$ We already know \(f'(u) = \frac{1}{2\sqrt{u}}\) and \(g'(x) = 4\), so we have: $$ \frac{dy}{dx}=\frac{1}{2\sqrt{g(x)}} \cdot 4 = \frac{1}{2\sqrt{4x+1}} \cdot 4. $$ Then, we can simplify this expression to obtain our answer: $$ \frac{dy}{dx} = \frac{4}{2\sqrt{4x+1}} = \frac{2}{\sqrt{4x+1}}. $$ So, the derivative of \(y=\sqrt{4x+1}\) is \(\frac{d}{dx}(y)=\frac{2}{\sqrt{4x+1}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
In calculus, the concept of a derivative is crucial as it represents the rate at which a function changes. For the function given in the exercise, specifically \(y = \sqrt{4x+1}\), the derivative helps us determine how swiftly the value of \(y\) changes as \(x\) changes.
The Chain Rule in differentiation is particularly useful when dealing with composite functions. A composite function, like \(y = \sqrt{4x+1}\), consists of an inner function \(g(x)\) and an outer function \(f(u)\). The Chain Rule states that in order to find the derivative of a composite function \(f(g(x))\), we calculate the derivative of the outer function with respect to the inner function, \(f'(g(x))\), and multiply it by the derivative of the inner function, \(g'(x)\).
This approach breaks down complex calculations into simpler parts, making it easier to manage and understand the derivatives of more intricate functions.
Inner and Outer Functions
When using the Chain Rule in calculus, it's important to identify the inner and outer functions accurately, as this sets the foundation for the rest of the calculation.
Let's break it down in the given exercise where \(y = \sqrt{4x+1}\). Here,
  • Inner Function: This is the piece of the composite function inside something else, which in this case is \(g(x) = 4x + 1\). The inner function is what gets substituted into the outer function.
  • Outer Function: This is the main operation surrounding the inner function, denoted as \(f(u) = \sqrt{u}\), where \(u = g(x)\).
Recognizing these functions helps us efficiently use the Chain Rule, by placing focus on how these parts interact with each other to form the overall composite function.
Simplifying Expressions
After applying the Chain Rule, the next critical step often involves simplifying the resulting expression. This makes the solution clearer and more concise.
In this particular exercise, we've applied the chain rule to find that the derivative of \(y = \sqrt{4x+1}\) is initially \(\frac{4}{2\sqrt{4x+1}}\). The expression at this stage can be cleaned up to make it more presentable. Simplifying involves reducing fractions and combining similar terms.
  • Here the fraction \(\frac{4}{2}\) simplifies to \(2\), making the expression \(\frac{2}{\sqrt{4x+1}}\).
This final simplified form is often sufficient for interpretation or for further mathematical applications, such as integration or solving differential equations. Keeping expressions simple is key to avoiding errors and ensuring that solutions are easy to understand and use.

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Most popular questions from this chapter

A mixing tank A 500 -liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min.}\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank is given by $$V(t)=500-0.5 t$$ a. Graph the mass function and verify that \(M(0)=0\) b. Graph the volume function and verify that the tank is empty when \(t=1000 \mathrm{min}\) c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) \(\underset{t \rightarrow 1000^{-}}{\operatorname{and}} C(t) ?\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\) e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\) f. For what times is the concentration of the solution increasing? Decreasing?

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$(x y+1)^{3}=x-y^{2}+8$$

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

Find \(d y / d x,\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}.\)

An equilateral triangle initially has sides of length \(20 \mathrm{ft}\) when each vertex moves toward the midpoint of the opposite side at a rate of \(1.5 \mathrm{ft} / \mathrm{min}\). Assuming the triangle remains equilateral, what is the rate of change of the area of the triangle at the instant the triangle disappears?
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