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Find the derivative of the following functions by first expanding the expression. Simplify your answers. $$h(x)=\sqrt{x}(\sqrt{x}-1)$$

Short Answer

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Question: Find the derivative of the function $$h(x) = \sqrt{x}(\sqrt{x} - 1)$$. Answer: The derivative of the given function is $$h'(x) = \frac{3}{2}\sqrt{x} - \frac{\sqrt{x}}{2x}$$.

Step by step solution

01

Expand and simplify the expression

First, we will expand the expression by multiplying \(\sqrt{x}\) with both terms inside the parentheses. After that, we will simplify the result. $$h(x) = \sqrt{x}(\sqrt{x}-1) = x\sqrt{x}-\sqrt{x}\cdot 1 =x\sqrt{x} - \sqrt{x} $$
02

Rewrite with rational exponents

Next, we will rewrite the expression using rational exponents so that it is easier to differentiate: $$h(x) = x\sqrt{x} - \sqrt{x} = x^{\frac{3}{2}} - x^{\frac{1}{2}}$$
03

Differentiate the simplified expression

Now, we will find the derivative of the rewritten function with respect to x using the power rule, which states that \(\frac{d}{dx}(x^n)=nx^{n-1}\): $$h'(x) = \frac{d}{dx}(x^{\frac{3}{2}} - x^{\frac{1}{2}}) = \frac{3}{2}x^{\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}$$
04

Rewrite the derivative with radical notation

Finally, we will rewrite the derivative using radical notation to obtain our final answer: $$h'(x) = \frac{3}{2}x^{\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}= \frac{3}{2}\sqrt{x}-\frac{1}{2x^{\frac{1}{2}}} =\frac{3}{2}\sqrt{x}-\frac{\sqrt{x}}{2x}$$ So the derivative of the given function is: $$h'(x) = \frac{3}{2}\sqrt{x}-\frac{\sqrt{x}}{2x}$$

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Key Concepts

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

Power Rule for Derivatives
The power rule is a fundamental technique in calculus for finding the derivative of functions of the form f(x) = x^n, where n is a real number. According to the power rule, the derivative of f(x) is calculated as f'(x) = nx^{n-1}. In essence, the exponent n is brought down in front as a coefficient, and the exponent of x is reduced by one. This rule greatly simplifies the derivative calculation process, making it possible to quickly find the rate of change or slope of a power function.

For example, if you have the function g(x) = x^5, applying the power rule, the derivative g'(x) would be 5x^4. This straightforward method extends to negative and fractional exponents, making it a versatile tool for various functions.
Rational Exponents
Rational exponents represent another way to express roots, such as square roots, cube roots, etc. In general, an expression like x^(m/n), where m and n are integers, is equivalent to the nth root of x raised to the mth power (nth root(x^m)). This form is very useful when differentiating because it allows us to apply the power rule directly.

For instance, the square root of x can be written as x^(1/2) and its derivative is simply found using the power rule: d/dx (x^(1/2)) = (1/2)x^(1/2 - 1) = (1/2)x^(-1/2). When functions involve roots, converting them to their rational exponent form simplifies the differentiation process.
Simplifying Expressions
Before applying the power rule or any rule of differentiation, it's often helpful to simplify the expression you're working with. Simplifying can include expanding products, factoring out common terms, canceling terms, and converting roots to exponent form. The goal is to transform the function into a form where the derivatives can be more easily and accurately computed.

For example, take the function h(x) = (x + 3)(x - 2). If we simplify by expanding the expression first, we get h(x) = x^2 + x - 6, a much more straightforward form for differentiating. Simplification is not just about making things easier; it's also about ensuring accuracy, as a less complex expression reduces the chances of making errors in the calculation.
Product Rule for Derivatives
The product rule is a derivative rule that allows us to differentiate functions that are products of two or more functions. It states that the derivative of a product of two functions u(x) and v(x) is given by u'v + uv', where u' and v' are the derivatives of u and v respectively. This rule is particularly important when you can't easily simplify the expression into a single term.

Let's consider the function f(x) = x^2 * sin(x). To differentiate this using the product rule, we find the derivatives separately: for x^2 (which is 2x) and for sin(x) (which is cos(x)), and combine them as per the rule to get f'(x) = 2x * sin(x) + x^2 * cos(x). The product rule is essential for handling more complex differentiation scenarios where two or more functions are being multiplied together.

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Most popular questions from this chapter

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general, $$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\)

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.

Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions \(\theta(t)\) and \(\varphi(t),\) respectively, where \(0 \leq t \leq 4\) and \(t\) is measured in minutes (see figure). These angles are measured in radians, where \(\theta=\varphi=0\) represent the starting position and \(\theta=\varphi=2 \pi\) represent the finish position. The angular velocities of the runners are \(\theta^{\prime}(t)\) and \(\varphi^{\prime}(t)\). a. Compare in words the angular velocity of the two runners and the progress of the race. b. Which runner has the greater average angular velocity? c. Who wins the race? d. Jean's position is given by \(\theta(t)=\pi t^{2} / 8 .\) What is her angular velocity at \(t=2\) and at what time is her angular velocity the greatest? e. Juan's position is given by \(\varphi(t)=\pi t(8-t) / 8 .\) What is his angular velocity at \(t=2\) and at what time is his angular velocity the greatest?

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(x^{\cos x}\right)$$

Logistic growth Scientists often use the logistic growth function \(P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{d}}}\) to model population growth, where \(P_{0}\) is the initial population at time \(t=0, K\) is the carrying capacity, and \(r_{0}\) is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. World population (part 1 ) The population of the world reached 6 billion in \(1999(t=0)\). Assume Earth's carrying capacity is 15 billion and the base growth rate is \(r_{0}=0.025\) per year. a. Write a logistic growth function for the world's population (in billions) and graph your equation on the interval \(0 \leq t \leq 200\) using a graphing utility. b. What will the population be in the year 2020? When will it reach 12 billion?

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