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Chapter 4: Q23MP (page 239)

Find $\frac{d}{d t} {鈭�}_{0}^{s i n t} \frac{s i n^{- 1} x}{x} d x$ .

Short Answer

Expert verified

The value of $\frac{d}{d t} {鈭�}_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x is t \cot t$ .

Step by step solution

01

Given Information

Given that . $\frac{d}{d t} {鈭�}_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x$ ...(1)

02

Formula Used

We know that $\frac{d}{d x} {鈭�}_{u (x)}^{v (x)} f (x, t) d t = f (x, v) \frac{d v}{d x} - f (x, u) \frac{d u}{d x} + {鈭�}_{u}^{v} \frac{鈭� f}{鈭� x} d t 鈭�$ ...(2)

03

Solve the given equation

Compare equation (1) and equation (2).

$\frac{d}{d t} {鈭�}_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = \frac{\sin^{- 1} (\sin t)}{\sin t} . \frac{d}{d t} (\sin t) - \frac{\sin^{- 1} (0)}{0} . \frac{d}{d t} (0) + {鈭�}_{0}^{\sin t} 0 d x$

Therefore,

$\frac{d}{d t} {鈭�}_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = \frac{t}{\sin t} . (\cos t) \frac{d}{d t} {鈭�}_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = t (\frac{\cos t}{\sin t}) \frac{d}{d t} {鈭�}_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = t \cot t$

Hence the value of $\frac{d}{d t} {鈭�}_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = t \cot t$ .

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Most popular questions from this chapter

To find the familiar "second derivative test "for a maximum or minimum point of the functions of two variables if $f_{x} = f_{y} = 0$ atlocalid="1664265078344" $(a, b)$ then,

localid="1664265157617" $(a, b)$ Is maximum point if at $(a, b), f_{x x} > 0, f_{y y} > 0 f_{x x} f_{y y} > f_{x y}^{2}$ .

$(a, b)$ Is maximum point if at $(a, b), f_{x x} < 0, f_{y y} < 0 f_{x x} f_{y y} > f_{x y}^{2}$

$(a, b)$ Is neither a maximum nor minimum point if $f_{x x} f_{y y} < f_{x y}^{2}$ .

If $P = r c o s t$ and role="math" localid="1664266535265" $r s i n t - 2 t e^{r} = 0$ , find $\frac{d P}{d t}$ .

Question: Show that satisfies $u (x, y) = \frac{y}{蟺} {鈭�}_{- 鈭�}^{鈭�} \frac{f (t) d t}{{(x - t)}^{2} + y^{2}} satisfies u_{x x} + u_{y y} = 0$ .

If $z = x^{2} + 2 y^{2}$ , $x = r c o s 胃$ , $y = r s i n 胃$ , find the following partial derivatives.

${(\frac{鈭� z}{鈭� y})}_{x}$ .

The acceleration of gravity can be found from the length land period T of a pendulum; the formula is $g = \frac{4 蟺^{2} l}{T^{2}}$ . Find the relative error ingin the worst case if the relative error inIis 5%, and the relative error inTis 2%.

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