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Chapter 10: Problem 836

If \(\mathrm{y}={ }^{\mathrm{x}} \sum_{\mathrm{r}=1} \tan ^{-1}\left[1 /\left(1+\mathrm{r}+\mathrm{r}^{2}\right)\right]\) then \((\mathrm{dy} / \mathrm{dx})=\) (a) \(\left[1 /\left(1+x^{2}\right)\right]\) (b) \(\left[1 /\left\\{1+(1+\mathrm{x})^{2}\right\\}\right]\) (c) 0 (d) \(\left[1 /\left\\{1-(\mathrm{x}+1)^{2}\right\\}\right]\)

Short Answer

Expert verified
(d) \(\frac{-1}{(x+1)^{2}-1}\)

Step by step solution

01

Understanding the summation notation

Rewrite the given function y to understand its structure. \(y(x) = \tan^{-1}\left[\frac{1}{1+1+1^{2}}\right] + \tan^{-1}\left[\frac{1}{1+2+2^{2}}\right] + ... + \tan^{-1}\left[\frac{1}{1+x+x^{2}}\right]\)
02

Differentiating the function y with respect to x

As the function is a sum of individual terms, differentiate each term separately. \(\frac{dy}{dx} = \frac{d}{dx} \left(\tan^{-1}\left[\frac{1}{1+1+1^{2}}\right] \right) + \frac{d}{dx} \left(\tan^{-1}\left[\frac{1}{1+2+2^{2}}\right] \right) + ... + \frac{d}{dx} \left(\tan^{-1}\left[\frac{1}{1+x+x^{2}}\right] \right) \)
03

Differentiating the arctangent function

Recall that the derivative of arctangent function, \(\tan^{-1}(u)\), is \(\frac{1}{1+u^2}\) times the derivative of u with respect to x. \(\frac{dy}{dx} = \frac{1}{(1+1+1^{2})^{2}} \cdot \frac{d}{dx} \left(\frac{1}{1+1+1^{2}}\right) + \frac{1}{(1+2+2^{2})^{2}} \cdot \frac{d}{dx} \left(\frac{1}{1+2+2^{2}}\right) + ... + \frac{1}{(1+x+x^{2})^{2}} \cdot \frac{d}{dx} \left(\frac{1}{1+x+x^{2}}\right) \)
04

Differentiating each term with respect to x

Now, we need to differentiate each term with respect to x. \(\frac{d}{dx} \left(\frac{1}{1+x+x^{2}}\right) = \frac{-d}{dx} \left(1+x+x^{2}\right) = -(1 + 2x)\)
05

Finalize the derivative dy/dx

Now, we can replace this derivative in our previous expression, obtaining the full derivative of y with respect to x. \(\frac{dy}{dx} = \frac{1}{(1+1+1^{2})^{2}} \cdot \frac{d}{dx} \left(\frac{1}{1+1+1^{2}}\right) + \frac{1}{(1+2+2^{2})^{2}} \cdot \frac{d}{dx} \left(\frac{1}{1+2+2^{2}}\right) + ... - (1+2x) \cdot \frac{1}{(1+x+x^{2})^{2}}\) This expression simplifies in terms of x only in its last term: \(\frac{dy}{dx} = ... + \frac{-(1 + 2x)}{(1 + x + x^{2})^{2}}\) Comparing this expression with the multiple-choice options provided, we can see that it matches option (d): \(\frac{dy}{dx} = \frac{-1}{(1+x+x^{2})^{2}}\) Thus, our answer is: (d) \(\left[ \frac{-1}{(x+1)^{2}-1} \right]\)

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Most popular questions from this chapter

If \(y=\sin (x / 2) \mid[1 /\\{\cos (x / 2) \cos x\\}]+[1 /\\{\cos x \cos (3 x / 2)\\}]\) \(+[1 /\\{\cos (3 \mathrm{x} / 2) \cos 2 \mathrm{x}\\}] \mid\) then \((\mathrm{dy} / \mathrm{dx})_{\mathrm{x}=(\pi / 2)}=\) (a) \((3 / 2)\) (b) \((1 / 2)\) (c) \(-1\) (d) 1

If \(\mathrm{e}^{\mathrm{y}}=\left(\mathrm{e}^{2} / \mathrm{x}^{2}\right)\) and \(\left[\left(\mathrm{d}^{2} \mathrm{y}\right) /\left(\mathrm{d} \mathrm{x}^{2}\right)\right]=\left(\mathrm{A} / \mathrm{x}^{2}\right)\) the \(\mathrm{A}=\) (a) \(-2\) (b) \((1 / 2)\) (c) 2 (d) \((1 / 3)\)

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) \((0,-3)\)

If \(\mathrm{y}=\tan ^{-1} \mid\left[\left\\{\sqrt{\left. \left.\left(1+\mathrm{x}^{2}\right)-\sqrt{(1-\mathrm{x}}^{2}\right)\right\\} /\left\\{\mathrm{V}\left(1+\mathrm{x}^{2}\right)+\sqrt{ \left.\left(1-\mathrm{x}^{2}\right)\right\\}}\right]}\right.\right.\) and \(z=\cos ^{-1} x^{2}\) then \((d y / d x)=\) (a) \((1 / 2)\) (b) \(\left[x /\left\\{\sqrt{ \left.\left(1-x^{4}\right)\right\\}}\right]\right.\) (c) \(\left(\mathrm{x}^{2} / 4\right)\) (d) \(\left(\mathrm{x}^{2} / 4\right)-(1 / 2)\)

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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