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Chapter 4: Q2P (page 201)

Given $w = \sqrt{u^{2} + v^{2}}$ ,
$u = \cos [Intan (p + \frac{1}{4} 蟺)] v = \sin [In \tan (p + \frac{1}{4} 蟺)]$ , find $\frac{d w}{d p}$ .

Short Answer

Expert verified

The value of $\frac{d w}{d p}$ is zero.

Step by step solution

01

Explanation of solution

The provided expressions are $w = \sqrt{u^{2} + v^{2}}$ ,

$u = \cos [Intan (p + \frac{1}{4} 蟺)] v = \sin [In \tan (p + \frac{1}{4} 蟺)]$

02

Chain Rule

The number of functions in the composition affects how many differentiation steps are required when using the chain rule to get the derivative of composite functions.

03

Calculation

Use the below equation.

$d w = \frac{鈭� w}{鈭� u} d u + \frac{鈭� w}{鈭� v} d v$

Therefore, the value of in term ofp is,

$w = \sqrt{[\cos {[In \tan (p + \frac{1}{4} 蟺)]}^{2} + \sin {[In \tan (p + \frac{1}{4} 蟺)]}^{2}]} = \sqrt{[\cos {[In \tan (p + \frac{1}{4} 蟺)]}^{2} + \sin^{2} [In \tan (p + \frac{1}{4} 蟺)]]}$

Use identity $\sin^{2} x + \cos^{2} x = 1$ in above equation.

$w = \sqrt{1} = 1$

Thus, w is a constant function, and that the derivative of a constant function is always zero.

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Most popular questions from this chapter

As in Problem 11, estimate $\sqrt[3]{(2.05)^{2} + (1.98)^{2}}$ .

Given that ${鈭�}_{0}^{鈭�} e^{- a x} s i n k x d x = \frac{k}{a^{2} + k^{2}}$ differentiate with respect to to show that ${鈭�}_{0}^{鈭�} x e^{- a x} s i n k x d x = \frac{2 k a}{{(a^{2} + k^{2})}^{2}}$ and differentiate with respect to $k$ to show that ${鈭�}_{0}^{鈭�} x e^{- a x} c o s k x d x = \frac{a^{2} - k^{2}}{{(a^{2} + k^{2})}^{2}}$ .

If $u + v = x^{3} - y^{3} + 4, u^{2} - v^{2} = x^{2} y^{2} + 1$ , find ${(\frac{鈭� u}{鈭� x})}_{y}, {(\frac{鈭� u}{鈭� x})}_{v}, {(\frac{鈭� x}{鈭� u})}_{y}, {(\frac{鈭� x}{鈭� u})}_{v}$ at $(x, y, u, v) = (2, - 1, 3, 2)$ .

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

${(\frac{鈭� z}{鈭� x})}_{y}$ .

About how much (in percent) does an error of 1% in x and y affect $x^{3} y^{2}$ ?

See all solutions

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