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Use the General Power Rule to find the derivative of the function. $$ f(x)=-3 \sqrt[4]{2-9 x} $$

Short Answer

Expert verified
The derivative of the function \(f(x)=-3 \sqrt[4]{2-9 x}\) is \(f'(x) = \frac{27}{4}(2-9x)^{-3/4}\).

Step by step solution

01

Rewriting the function

Rewrite the function \(f(x)= -3 \sqrt[4]{2-9 x}\) in the form that suits the power rule for derivatives. Express the fourth root as a power of \(1/4\). So the function becomes \(f(x)=-3 (2-9 x)^{1/4}\).
02

Identify u

Identify the function u and compute its derivative, i.e., \(u = 2-9x\) and find its derivative \(u' = -9\).
03

Applying the rule

Apply the General power rule \((u^n)' = n*u^{n-1}*u'\), so the derivative of \(f(x)\) will be \(f'(x) = -3*1/4*(2-9x)^{-3/4}*(-9)\).
04

Simplify

Simplify the function to get \(f'(x) = \frac{27}{4}(2-9x)^{-3/4}\).

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Key Concepts

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

Derivative of Functions
Derivatives are essential in calculus. They provide us with a way to understand how a function changes at any point. In simple terms, when we find the derivative of a function, we're discovering the rate at which the function values change with respect to changes in the variable. For instance, if you have a position vs. time graph, the derivative would represent the speed, or how fast the position changes over time.
The notation for a derivative is often written as \(f'(x)\), \(dy/dx\), or \(Df(x)\), where \(f(x)\) is the original function. Derivatives are used in various fields, from physics to engineering, to understand dynamic systems and predict future behavior based on present state changes. When dealing with complex functions, finding derivatives becomes crucial for optimization and analysis. It's important to break down a function into simpler parts if possible, and apply rules like the Chain Rule or Power Rule to differentiate each part correctly.
Power Rule for Derivatives
The Power Rule is one of the most fundamental rules in calculus used to find derivatives. It is useful for functions of the form \(x^n\), where \(n\) is any real number. The Power Rule states that if \(f(x) = x^n\), the derivative \(f'(x)\) is \(nx^{n-1}\).
To use this rule effectively, you must ensure the function is expressed in a suitable form. For example, a fourth root like \(\sqrt[4]{a}\) should be written as \(a^{1/4}\) so the Power Rule can be applied. *General Power Rule* is a variation that helps with functions in the form \(u^n\), where \(u\) is not just \(x\). It's given by the formula \((u^n)' = n \cdot u^{n-1} \cdot u'\).
This rule is incredibly powerful for efficiently finding derivatives without having to expand polynomials or manually apply multiple differentiation steps, especially for complex composite functions. The General Power Rule is what was used in the original exercise to differentiate the function after rewriting it.
Simplifying Derivatives
Simplifying derivatives is an important skill to make them easier to understand and use. Once you've applied a rule like the Power Rule, you'll often end up with an expression that can be cleaned up or reduced. Simplifying involves combining like terms, multiplying constants, and rearranging terms into a more approachable expression.
In the given exercise, after applying the General Power Rule, the expression for the derivative \(f'(x)\) was \(-3 \cdot \frac{1}{4} \cdot (2-9x)^{-3/4} \cdot (-9)\). This solution is then simplified to become \(f'(x) = \frac{27}{4} (2-9x)^{-3/4}\).
This process involves recognizing common factors, performing arithmetic on coefficients, and ensuring that the final expression represents the same mathematical quantity as the derived form. Simplifying makes the function much clearer and ready to be used for further analysis or application, such as evaluating it at certain points or integrating it later.

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Most popular questions from this chapter

Medicine The effectiveness \(E\) (on a scale from 0 to 1 ) of a pain-killing drug \(t\) hours after entering the bloodstream is given by $$ E=\frac{1}{27}\left(9 t+3 t^{2}-t^{3}\right), \quad 0 \leq t \leq 4.5 $$ Find the average rate of change of \(E\) on each indicated interval and compare this rate with the instantaneous rates of change at the endpoints of the interval. $$ \begin{array}{llll}{\text { (a) }[0,1]} & {\text { (b) }[1,2]} & {\text { (c) }[2,3]} & {\text { (d) }[3,4]}\end{array} $$

(a)Find an equation of the tangent line to the graph of the function at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. $$ y=x^{3}+x \quad(-1,-2) $$

Marginal Profit When the price of a glass of lemonade at a lemonade stand was \(\$ 1.75,400\) glasses were sold. When the price was lowered to \(\$ 1.50,500\) glasses were sold. Assume that the demand function is linear and that the variable and fixed costs are \(\$ 0.10\) and \(\$ 25\), respectively. (a) Find the profit \(P\) as a function of \(x,\) the number of glasses of lemonade sold. (b) Use a graphing utility to graph \(P,\) and comment about the slopes of \(P\) when \(x=300\) and when \(x=700\). (c) Find the marginal profits when 300 glasses of lemonade are sold and when 700 glasses of lemonade are sold.

Find \(f^{\prime}(x)\) $$ f(x)=x\left(x^{2}+1\right) $$

Marginal Profit The profit \(P(\text { in dollars) from selling } x\) units of a product is given by $$P=36,000+2048 \sqrt{x}-\frac{1}{8 x^{2}}, \quad 150 \leq x \leq 275$$ Find the marginal profit for each of the following sales. $$ \begin{array}{ll}{\text { (a) } x=150} & {\text { (b) } x=175 \quad \text { (c) } x=200} \\ {\text { (d) } x=225} & {\text { (e) } x=250 \quad \text { (f) } x=275}\end{array} $$

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