/*! 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 26 \(f(x)=x \sqrt{4 a x-x^{2}}\) ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 26

\(f(x)=x \sqrt{4 a x-x^{2}}\) \(f^{\prime}(x)=\sqrt{4 a x-x^{2}}+\frac{x(2 a-x)}{\sqrt{4 a x-x^{2}}}\) \(=\frac{6 a x-2 x^{2}}{\sqrt{4 a x-x^{2}}}=\frac{2 x(3 a-x)}{\sqrt{x(4 a-x)}}\)

Short Answer

Expert verified
Question: Verify if the given derivative of the function \(f(x)=x \sqrt{4 a x-x^2}\) is correct, where the derivative is provided as \(f^{\prime}(x)=\frac{6 a x-2 x^2}{\sqrt{4 a x-x^2}}\). Answer: Yes, the given derivative is correct.

Step by step solution

01

Identify the functions to apply the product rule

We have the function \(f(x) = x \sqrt{4 a x-x^2}\). To use the product rule for differentiation, we must recognize the two separate functions being multiplied together: Function 1: \(g(x) = x\) Function 2: \(h(x) = \sqrt{4 a x-x^2}\) The product rule states that \((g(x)h(x))^{\prime} = g^{\prime}(x)h(x) + g(x)h^{\prime}(x)\).
02

Differentiate Function 1 (g(x))

We will now differentiate \(g(x) = x\) with respect to \(x\): \(g^{\prime}(x) = \frac{d}{dx}(x) = 1\)
03

Differentiate Function 2 (h(x))

We will now differentiate \(h(x) = \sqrt{4 a x-x^2}\) with respect to \(x\): First, rewrite the square root as a power of one-half: \(h(x) = (4 a x-x^2)^{\frac{1}{2}}\) Next, apply the chain rule: \(h^{\prime}(x) = \frac{1}{2}(4 a x-x^2)^{-\frac{1}{2}}\cdot\frac{d}{dx}(4 a x-x^2)\) Differentiate the function inside the parenthesis: \(\frac{d}{dx}(4 a x-x^2) = 4a-2x\) Now plug this into the previous equation to complete the differentiation of h(x): \(h^{\prime}(x) = \frac{1}{2}(4 a x-x^2)^{-\frac{1}{2}}(4a-2x) = \frac{4a-2x}{2\sqrt{4 a x-x^2}}\)
04

Apply the product rule

We will now apply the product rule, plugging in the derivatives we found in Steps 2 and 3 and the original functions g(x) and h(x): \(f^{\prime}(x) = g^{\prime}(x)h(x) + g(x)h^{\prime}(x) = 1 \cdot \sqrt{4 a x-x^2} + x \cdot \frac{4a-2x}{2\sqrt{4 a x-x^2}}\)
05

Simplify the expression

Simplify the expression: \(f^{\prime}(x) = \frac{\sqrt{4 a x-x^2}}{1} + \frac{x(4a-2x)}{2\sqrt{4 a x-x^2}}\) To have a common denominator, multiply the first term by \(2\) on the top and bottom: \(f^{\prime}(x) = \frac{2\sqrt{4 a x-x^2}}{2} + \frac{x(4a-2x)}{2\sqrt{4 a x-x^2}}\) Now add the two terms together: \(f^{\prime}(x) = \frac{2\sqrt{4 a x-x^2} + x(4a-2x)}{2\sqrt{4 a x-x^2}}\) Factor out \(2x\) from the numerator: \(f^{\prime}(x) = \frac{2x(3a-x)}{2\sqrt{4 a x-x^2}}\) Finally, simplify by cancelling out the \(2\)s: \(f^{\prime}(x) = \frac{x(3a-x)}{\sqrt{4 a x-x^2}}\) #Solution# We found the derivative of the function \(f(x) = x \sqrt{4 a x-x^2}\) to be \(f^{\prime}(x) = \frac{x(3a-x)}{\sqrt{4 a x-x^2}}\), which matches the given derivative in the exercise. This verifies the correctness of the given derivative.

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Most popular questions from this chapter

\(f(x)=x^{3}+4 x^{2}+\lambda x+1\) \(f^{\prime}(x)=3 x^{2}+8 x+\lambda<0\) roots are \(-2 \& \frac{-2}{3}\) Product \(=+\frac{4}{3}\) \(\Rightarrow \lambda=+4\) Hence, A is correct

$\mathrm{h}(\mathrm{x})=\mathrm{f} \circ \mathrm{g}(\mathrm{x}), \quad \mathrm{h}(0)=0$ If \(x\) increase, \(h(x)\) decreases As $p(x)=h\left(x^{3}-2 x^{2}+2 x\right)-h(4) \forall x \in(0,2]$ as \(x \rightarrow 0\) \(p(0)=h\left(0^{3}-2 \cdot 0^{2}+2 \cdot 0\right)-h(4)\) \(=0-h(4)\) \(=-h(4)\) As \(\mathrm{p}(\mathrm{x})\) is decreasing function. $\Rightarrow p(2)=h\left((2)^{3}-2.2^{2}+2.2\right)-h(4)$ $\begin{aligned} &=\mathrm{h}(4)-\mathrm{h}(4) \\ &=0 \\ \Rightarrow \mathrm{p}(\mathrm{x}) & \in(0,-\mathrm{h}(4)) \end{aligned}$ Hence, A is correct

\(f(x)=g(x)\) \(\sin 3 x+\cos x=\cos 3 x+\sin x\) \(\sin 3 x-\sin x=\cos 3 x-\cos x\) \(\cos 2 x \sin x=-\sin 2 x \sin x\) \(\Rightarrow \sin x(\cos 2 x+\sin 2 x)=0\) \(\Rightarrow \sin x \sin \left(\frac{\pi}{4}+2 x\right)=0\) \(\Rightarrow\) either \(\sin x=0\) or \(\sin \left(\frac{\pi}{4}+2 x\right)=0\) \(x=0, \pi, \frac{3 \pi}{8}, \frac{7 \pi}{8}\) Hence, \(\mathrm{C}\) is correct

\(x=t^{2} \quad \& y=3 t+t^{3}\) \(\frac{d x}{d t}=2 t \quad \& \quad \frac{d y}{d t}=3+3 t^{2}\) \(\frac{d y}{d x}=\frac{3}{2} \frac{\left(1+t^{2}\right)}{t}\) $\frac{d^{2} y}{d x^{2}}=\frac{3}{2} \frac{t(2 t)-\left(1+t^{2}\right)}{t^{2}} \times \frac{1}{2 t}$ \(=\frac{3}{4} \frac{\left(t^{2}-1\right)}{t^{3}}\) $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}} \rightarrow \frac{-}{ }_{-1}^{+} \underset{0}{1} \stackrel{+}{ }_{1}^{+}$ Hence, \(\mathrm{C}\) is correct

\(g^{\prime}(x)=2\left(6 x^{2}-6 x\right)\) $f^{\prime}\left(2 x^{3}-3 x^{2}\right)+\left(12 x-12 x^{2}\right)+f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)$ $g^{\prime}(x)=\left(12 x^{2}-12 x\right)\left[f^{\prime}\left(2 x^{3}-3 x^{2}\right)-f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)\right]$ $=12 x(x-1)\left[f^{\prime}\left(2 x^{3}-3 x^{2}\right)-f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)\right]$ For increasing, If \(x \in(1, \infty)\) \(2 x^{3}-3 x^{2}>6 x^{2}-4 x^{3}-3\) \(6 x^{3}-9 x^{2}+3>0\) \(2 x^{3}-3 x^{2}+1>0\) \(\mathrm{x} \in(1, \infty)\) If \(\mathrm{x} \in(0,1)\) \(6 x^{2}-4 x^{3}-3<2 x^{3}-3 x^{2}\) \(2 x^{3}-3 x^{2}+1>0\) no soln If \(x<0\) \(2 x^{3}-3 x^{2}+1>0\) \(x \in\left(-\frac{1}{2}, 0\right)\) Hence, B is correct
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