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Chapter 2: Problem 3

A function \(\mathrm{f}\), defined for all positive real numbers, satisfies the equation \(\mathrm{f}\left(\mathrm{x}^{2}\right)=\mathrm{x}^{3}\) for every \(x>0\). Determine \(\mathbf{f}^{\prime}(4)\).

Short Answer

Expert verified
Answer: The value of the derivative of the function \(f\) at the point \(x=4\) is 3.

Step by step solution

01

Calculate the derivative of both sides of the equation using the chain rule

Start by differentiating the given equation \(f(x^2) = x^3\) with respect to \(x\). To do this, we need to apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function: \(\frac{d}{dx}[f(x^2)] = \frac{d}{dx}[x^3]\) Now, applying the chain rule: \(f^{\prime}(x^2) \cdot 2x = 3x^2\)
02

Determine the input of the function f to calculate \(f^{\prime}(4)\)

To find \(f^{\prime}(4)\), we need to find \(f^{\prime}(y)\) with \(y = x^2\) such that \(y = 4\). Let's find the value for \(x\): \(x^2 = 4 \Rightarrow x = \sqrt{4} = 2\) Thus, we have: \(f^{\prime}(4) = f^{\prime}(2^2)\)
03

Plug the value of x into the chain rule equation

Now, substitute \(x = 2\) into the chain rule equation: \(f^{\prime}(4) \cdot 2(2) = 3(2)^2 \implies f^{\prime}(4) \cdot 4 = 12\)
04

Solve for \(f^{\prime}(4)\) and find the value

Finally, solve for \(f^{\prime}(4)\): \(f^{\prime}(4) = \frac{12}{4}\) \(f^{\prime}(4) = 3\) So, therefore, \(f^{\prime}(4) = 3\).

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Most popular questions from this chapter

Prove that if \(|x|<1\) \(\frac{x^{3}}{1.3}-\frac{x^{5}}{3.5}+\frac{x^{7}}{5.7}-\ldots=\frac{1}{2}\left(1+x^{2}\right) \tan ^{-1} x-\frac{1}{2} x\)

Suppose that the velocity function of a particle moving along a line is \(v(t)=3 t^{3}+2\). Find the average velocity of the particle over the time interval \(1 \leq \mathrm{t} \leq 4\) by integrating.

(a) Make a conjecture about the value of the limit \(\lim _{k \rightarrow 0} \int_{1}^{b} t^{k-1} d t(b>0)\) (b) Check your conjecture by evaluating the integral and finding the limit. [Hint: Interpret the limit as the definition of the derivative of an exponential function]

(a) Show that \(1 \leq \sqrt{1+x^{3}} \leq 1+x^{3}\) for \(x \geq 0\)(b) Show that \(1 \leq \int_{0}^{1} \sqrt{1+x^{3}} d x \leq 1.25\).

Prove the inequalities: (i) \(\int_{1}^{3} \sqrt{x^{4}+1} d x \geq \frac{26}{3}\)(iii) \(\frac{1}{17} \leq \int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx} \leq \frac{7}{24}\).
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