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Chapter 4: Q27MP (page 239)

Show that $\frac{d}{d x} {鈭�}_{c o s x}^{s i n x} \sqrt{1 - t^{2}} d t = 1$ .

Short Answer

Expert verified

It is proved that $\frac{d}{d x} {鈭�}_{\cos x}^{\sin x} \sqrt{1 - t^{2}} d t = 1$ .

Step by step solution

01

Given Information

Given that . $\frac{d}{d x} {鈭�}_{\cos x}^{\sin x} \sqrt{1 - t^{2}} d t$ ...(1)

02

Formula Used

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

03

Solving the given equation

Compare equation (1) and equation (2).

\frac{d}{d x} {鈭�}_{\cos x}^{\sin x} \sqrt{1 - t^{2}} d t = \sqrt{1 - \sin^{2} x} . \frac{d}{d x} (\sin x) - \sqrt{1 - \cos^{2} x} . \frac{d}{d x} (\cos x) + {鈭�}_{\cos x}^{\sin x} 0 . d t \frac{d}{d x} {鈭�}_{\cos x}^{\sin x} \sqrt{1 - t^{2}} d t = \sqrt{\cos^{2} x} . (\cos x) - \sqrt{\sin^{2} x} . (- \sin x) \frac{d}{d x} {鈭�}_{\cos x}^{\sin x} \sqrt{1 - t^{2}} d t = (\cos x) . (\cos x) - (\sin x) . (- \sin x) \frac{d}{d x} {鈭�}_{\cos x}^{\sin x} \sqrt{1 - t^{2}} d t = \cos^{2} x + \sin^{2} x \frac{d}{d x} {鈭�}_{\cos x}^{\sin x} \sqrt{1 - t^{2}} d t = 1

Hence Proved

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Most popular questions from this chapter

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

${(\frac{鈭� z}{鈭� x})}_{胃}$ .

Given $f (x, y, z) = 0$ and $g (x, y, z) = 0,$ find a formula for $d y / d x$ .

Find by the Lagrange multiplier method the largest value of the product of three positive numbers if their sum is 1.

Find the two-variable Maclaurin series for the following functions.

$\sqrt{1 + xy}$

To find the familiar "second derivative test "for a maximum or minimum point. That is show that $f^{'} (a) = 0$ , then $f^{''} (a) > 0$ implies a minimum point at $x = a$ and $f^{''} (a) < 0$ implies a maximum point at $x = a$ .

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