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Chapter 4: Q 7P (page 201)

Given c = sin(a - b ), $b = a e^{2 a}$ , find $\frac{dc}{db}$ .

Short Answer

Expert verified

The $\frac{d c}{d b}$ is $\cos (a - b) (1 - 2 b - e^{2 a})$ .

Step by step solution

01

Explanation of solution

The provided expressions are $c = \sin (a - b), b = a e^{2 a}$ .

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

Find $\frac{d b}{d a}$ .

$\frac{d b}{d a} = a e^{2 a} (2) + e^{2 a} = 2 a e^{2 a} + e^{2 a}$

Find .

$\frac{d c}{d a} = \cos (a - b) (1 - \frac{d b}{d a}) = \cos (a - b) (1 - 2 a e^{2 a} - e^{2 a}) \frac{d c}{d a} = \cos (a - b) (1 - 2 b - e^{2 a})$

Hence the $\frac{d c}{d a}$ is $\cos (a - b) (1 - 2 b - e^{2 a})$ .

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Most popular questions from this chapter

Question: Show that satisfies $u (x, y) = \frac{y}{π} \int_{- \infty}^{\infty} \frac{f (t) d t}{{(x - t)}^{2} + y^{2}} satisfies u_{x x} + u_{y y} = 0$ .

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

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

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

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

If $u = f (x - c t) + g (x + c t)$ , show that $\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} u}{\partial t^{2}} .$

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

$\frac{\partial^{2} z}{\partial r \partial x}$ .

See all solutions

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