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Chapter 10: Problem 854

If \(y=(x \log x)^{\log \log x}\) for \((d y / d x)=A^{\log \log x-1}(B+(\log x+2) \log\) \(\log x\) ) then \(A B=\) (a) \(x \log x\) (b) \(x(\log x)^{2}\) (c) \(x^{2} \cdot \log x\) (d) \(\log x\)

Short Answer

Expert verified
The value of AB is \(x(\log x)^{2}\), which corresponds to option (b).

Step by step solution

01

Differentiate y with respect to x

To find the derivative \(\frac{dy}{dx}\), we will apply both the chain rule and the product rule. The chain rule states that if we have a composite function, then its derivative is the product of the derivative of the outer function applied to the inner function and the derivative of the inner function. Here, our outer function is \(u^{\log \log x}\) and our inner function is \(u = x \log x\). Using the chain rule, \[\frac{dy}{dx}=\frac{d}{dx}\Big((x \log x)^{\log \log x}\Big) = (\log \log x)(x \log x)^{\log \log x -1} \frac{d}{dx}(x \log x)\]
02

Differentiate the inner function with respect to x

The inner function is \(u = x \log x\). To differentiate this, we will use the product rule, which states that the derivative of the product of two functions is the sum of the product of the first function and the derivative of the second function and the product of the second function and the derivative of the first function. Using the product rule, we have: \[\frac{d}{dx}(x \log x) = x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(x)\] Now differentiate each term: \[\frac{d}{dx}(x \log x) = x \frac{1}{x} + \log x (1) = 1 + \log x\]
03

Plug ∂(x ∙ log(x))/∂x back into dy/dx

Now substitute back the value of \(\frac{d}{dx}(x \log x)\) in the derivative of the main function: \[\frac{dy}{dx} = (\log \log x)(x \log x)^{\log \log x -1} (1 + \log x)\]
04

Compare with given derivative to find A and B

Given derivative is: \(\frac{dy}{dx} = A^{\log \log x - 1}(B+(\log x + 2)\log \log x)\). By comparing it with the derivative we found above, we can equate the two expressions: \[(\log \log x)(x \log x)^{\log \log x -1} (1 + \log x) = A^{\log \log x - 1}(B+(\log x + 2)\log \log x)\] Now comparing the two sides, we can conclude that: \(A = (x \log x)\) and \(B = (\log x + 2) \log \log x - \log x\)
05

Find the value of AB

To find the value of AB, we just multiply A and B. \[AB = (x \log x)\Big((\log x + 2) \log \log x - \log x\Big)\] Looking for a matching option, we find that this expression corresponds to option (b), which is \(x(\log x)^{2}\). Therefore, the answer is (b).

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Most popular questions from this chapter

The point on the curve \(\mathrm{y}=(\mathrm{x}-2)(\mathrm{x}-3)\) at which the tangent makes an angle of \(225^{\circ}\) with positive direction of \(\mathrm{x}\) -axis has co-ordinates (a) \((0,3)\) (b) \((3,0)\) (c) \((-3,0)\) (d) \((0,-3)\)

For every \(\mathrm{x}, \mathrm{x} \in \mathrm{R}, \mathrm{f}(\mathrm{x})=(\mathrm{a}+2) \mathrm{x}^{3}-3 \mathrm{ax}^{2}+9 \mathrm{ax}-1\) the function is decreasing then a (a) \((-4,-3)\) (b) \(\overline{(-3,-2)}\) (c) \((3,0)\) (d) \((-1,-3)\)

If \(x^{2} e^{y}+2 x y e^{x}+23=0\) then \((d y / d x)=\) (a) \(2 \mathrm{xe}^{\mathrm{y}-\mathrm{x}}+2 \mathrm{y}(\mathrm{x}+1)\) (b) \(2 \mathrm{xe}^{\mathrm{x}-\mathrm{y}}-3 \mathrm{y}(\mathrm{x}+1)\) (c) \(\left[\left\\{-2 x e^{y}-e^{x} \cdot 2 y(x+1)\right\\} /\left\\{x\left(x e^{y}+e^{x} \cdot 2\right)\right\\}\right]\) (d) \(2 \mathrm{xe}^{\mathrm{y}-\mathrm{x}}-\mathrm{y}(\mathrm{x}+1)\)

If \(\mathrm{y}=\mathrm{x}^{\mathrm{x}}\) and \(\left[\left(\mathrm{d}^{2} \mathrm{y}\right) /\left(\mathrm{d} \mathrm{x}^{2}\right)\right]-(\mathrm{y} / \mathrm{x})=(1 / \alpha) \cdot(\mathrm{dy} / \mathrm{d} \mathrm{x})^{2}\) then (a) \(\mathrm{x}^{\mathrm{y}}\) (b) \(x^{x}\) (c) \(\mathrm{y}^{\mathrm{x}}\) (d) \(x\)

If \(\mathrm{f}^{\prime}(\mathrm{x})>0\) and \(\mathrm{g}^{\prime}(\mathrm{x})<0 \mathrm{x} \in \mathrm{R}\) then (a) \(\mathrm{f}(\mathrm{g}(\mathrm{x}))>\mathrm{f}(\mathrm{g}(\mathrm{x}+1))\) (b) \(f(g(x))g(f(x+1))\) (d) \(\mathrm{g}(\mathrm{f}(\mathrm{x}))>\mathrm{g}(\mathrm{f}(\mathrm{x}-1))\)
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