/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Find the derivatives of the give... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivatives of the given functions. Assume that \(a, b, c,\) and \(k\) are constants. $$y=\sqrt{x}$$

Short Answer

Expert verified
The derivative is \( \frac{1}{2\sqrt{x}} \).

Step by step solution

01

Recognize the Function Form

The given function is \( y = \sqrt{x} \). This can be rewritten as \( y = x^{1/2} \), which is a power function, allowing us to apply the power rule for derivatives.
02

Apply the Power Rule

The power rule states that the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \). Applying this to \( y = x^{1/2} \), we get:\[ \frac{dy}{dx} = \frac{1}{2}x^{-1/2} \].
03

Simplify the Derivative

The expression \( \frac{1}{2}x^{-1/2} \) can be simplified further. Since \( x^{-1/2} \) is the same as \( \frac{1}{\sqrt{x}} \), we can write the derivative as:\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \].

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Key Concepts

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

Power Rule
The power rule is a fundamental technique in calculus for finding the derivative of power functions. It is particularly useful for functions in the form of \( y = x^n \), where \( n \) is any real number. According to the power rule, the derivative of \( x^n \) with respect to \( x \) is given by the formula:\[ \frac{d}{dx}x^n = nx^{n-1} \]This means you multiply the power \( n \) by the term \( x^{n-1} \). Let's break it down further:- Bring down the power as a coefficient.- Reduce the power by one.For example, if \( y = x^{3} \), applying the power rule gives \( \frac{dy}{dx} = 3x^{2} \). This procedure is quick and handy for simplifying and differentiating polynomial expressions. When applied to \( y = x^{1/2} \), as in our original problem, we get the derivative \( \frac{dy}{dx} = \frac{1}{2}x^{-1/2} \). Remember, understanding this rule helps you tackle many types of derivative problems efficiently.
Derivative of Square Root
The derivative of a square root function can be initially approached by rewriting it in a form where the power rule can be applied. The function \( y = \sqrt{x} \) can be rewritten as \( y = x^{1/2} \). This step transforms the square root into a power expression, enabling us to use the powerful tool of calculus, the power rule.1. Convert square roots into exponent form. - Replace \( \sqrt{x} \) with \( x^{1/2} \). 2. Apply the power rule to find the derivative. - The power rule states: \( \frac{d}{dx}x^n = nx^{n-1} \). - For \( y = x^{1/2} \), we calculate: \( \frac{dy}{dx} = \frac{1}{2}x^{-1/2} \).By understanding how to work with square roots, and expressing them as exponents, you open up a straightforward path to finding derivatives using familiar rules.
Simplifying Expressions
Simplifying expressions in calculus, especially derivatives, is an essential step to reach a more manageable and easy-to-understand form. Starting from where the derivative was found as \( \frac{dy}{dx} = \frac{1}{2}x^{-1/2} \), simplification involves expressing it in terms of positive exponents or more easily interpretable forms.Here's how we simplify:- Recognize that negative exponents indicate reciprocals.- Rewrite \( x^{-1/2} \) as \( \frac{1}{\sqrt{x}} \).So, this expression becomes:\[ \frac{dy}{dx} = \frac{1}{2} \times \frac{1}{\sqrt{x}} = \frac{1}{2\sqrt{x}} \]This simplification highlights the importance of transforming derivatives into forms that are easier to integrate into further problems or applications. Mastering how to manipulate and simplify mathematical expressions is crucial for success in calculus.

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Most popular questions from this chapter

Let \(f\) be a differentiable function and let \(L\), be the linear function \(L(x)=f(a)+k(x-a)\) for some constant \(a\) Decide whether the following statements are true or false for all constants \(k .\) Explain your answer. (a) \(L\) is the local linearization for \(f\) near \(x=a\) (b) If \(\lim _{x \rightarrow a}(f(x)-L(x))=0,\) then \(L\) is the local linearization for \(f\) near \(x=a\).

Derive the chain rule using local linearization. THint: In other words, differentiate \(f(g(x)),\) using \(g(x+h) \approx\) \(\left.g(x)+g^{\prime}(x) h \text { and } f(z+k) \approx f(z)+f^{\prime}(z) k .\right\rfloor\)

The following problem is a generalization of Problem 31 . Show that any function of the form $$y=A e^{x}+B e^{-x}, \quad A>0, B>0$$can be written, for some \(K\) and \(c,\) in the form $$y=K \cosh (x-c)$$What does this tell you about the graph of \(y=A e^{x}+\) \(B e^{-x} ?\)

Find a formula for the error \(E(x)\) in the tangent line approximation to the function near \(x=a\). Using a table of values for \(E(x) /(x-a)\) near \(x=a\), find a value of \(k\) such that \(E(x) /(x-a) \approx k(x-a) .\) Check that, approximately, \(k=f^{\prime \prime}(a) / 2\) and that \(E(x) \approx\left(f^{\prime \prime}(a) / 2\right)(x-a)^{2}\). $$f(x)=\cos x, a=0$$

Suppose that \(f^{\prime \prime}\) and \(g^{\prime \prime}\) exist and that \(f\) and \(g\) are concave up for all \(x .\) Are the statements in Problems \(91-92\) true or false for all such \(f\) and \(g\) ? If a statement is true, explain how you know. If a statement is false, give a counterexample. \(f(x)+g(x)\) is concave up for all \(x\)

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