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Chapter 4: Q4P (page 201)

Given $x = In (u^{2}, v^{2}), u = t^{2}$ and $v = \cos t . find \frac{dx}{dt}$ .

Short Answer

Expert verified

The value of $\frac{d x}{d t}$ is $\frac{4 u t + 2 v \sin t}{u^{2} - v^{2}}$ .

Step by step solution

01

Explanation of solution

The provided expressions are $x = In (u^{2} - v^{2}), u = t^{2} and v = cost$ .

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

Consider the below equation to find $d x$ .

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

Find $\frac{鈭� x}{鈭� u}$ .

$\frac{鈭� x}{鈭� u} = \frac{1}{u^{2} - v^{2}} 2 u = \frac{2 u}{u^{2} - v^{2}}$

Find $\frac{鈭� x}{鈭� v}$ .

$\frac{鈭� x}{鈭� v} = \frac{1}{u^{2} - v^{2}} (- 2 v) = \frac{- 2 v}{u^{2} - v^{2}}$

Find dx.

$d x = \frac{- 2 v}{u^{2} - v^{2}} d u - (\frac{2 u}{u^{2} - v^{2}}) d v$

Use the value of $d u = 2 d t$ and $d v = - \sin t d t$ also differentiate above equation with respect to t.

$\frac{d x}{d t} = \frac{2 v}{u^{2} - v^{2}} (2 t) - (\frac{2 v}{u^{2} - v^{2}}) (- \sin t) = \frac{4 u t + 2 v \sin t}{u^{2} - v^{2}}$

Hence the value of $\frac{d x}{d t}$ is $\frac{4 u t + 2 v \sin t}{u^{2} - v^{2}}$ .

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Most popular questions from this chapter

Use differentials to show that, for large n and small a, $\sqrt{n + a} - \sqrt{n} 鈮� \frac{a}{2 \sqrt{n}}$ .Find the approximate value of $\sqrt{10^{26} + 5} - \sqrt{10^{26}}$ .

Use differentials to show that, for very large n , $\frac{1}{{(n + 1)}^{3}} - \frac{1}{n^{3}} 鈮� \frac{3}{n^{4}}$ .

For, $u = e^{x} c o s y$

(a) Verify that ${鈭�}^{2} u / 鈭� x 鈭� y = {鈭�}^{2} u / 鈭� y 鈭� x$ ;

(b) Verify that ${鈭�}^{2} u / 鈭� x^{2} = {鈭�}^{2} u / 鈭� y^{2} = 0$

Question: If ${鈭�}_{u}^{v} e^{- t^{2}} dt = x and u^{v} = y, find {(\frac{鈭� u}{鈭� x})}_{y}, {(\frac{鈭� u}{鈭� y})}_{x}, and {(\frac{鈭� y}{鈭� x})}_{u} at u = 2, v = 0$

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

${(\frac{鈭� z}{鈭� x})}_{r}$ $(\frac{鈭� z}{鈭� x})$ .

See all solutions

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