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Chapter 4: Q5P (page 201)

If we are given $z = z (x, y)$ and $y = y (x)$ , show that the chain rule (5.1) gives $\frac{d z}{d x} = \frac{鈭� z}{鈭� x} + \frac{鈭� z}{鈭� y} \frac{d y}{d x}$ $\frac{d z}{d x} = \frac{鈭� z}{鈭� x} + \frac{鈭� z}{鈭� y} \frac{d y}{d x}$

Short Answer

Expert verified

The chain rule $\frac{d z}{d x} = \frac{鈭� z}{鈭� x} + \frac{鈭� z}{鈭� y} \frac{d y}{d x}$ is proved.

Step by step solution

01

Explanation of solution

The provided expressions are $y = y (x)$ and $z = z (x, y)$ .

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 relation $y = y (x)$

Here, z can be expressed as,

$z = z (x, y (x)) = z (x)$

Therefore, $\frac{d z}{d x}$ exists.

Apply the chain rule.

$\frac{d z}{d x} = \frac{鈭� z}{鈭� x} \frac{d x}{d t} + \frac{鈭� z}{鈭� y} \frac{d y}{d t} - - - (1)$

Substitute in equation (1).

$\frac{d z}{d x} = \frac{鈭� z}{鈭� x} \frac{d x}{d x} + \frac{鈭� z}{鈭� y} \frac{d y}{d x} = \frac{鈭� z}{鈭� x} (1) + \frac{鈭� z}{鈭� y} \frac{d y}{d x} = \frac{鈭� z}{鈭� x} + \frac{鈭� z}{鈭� y} \frac{d y}{d x}$

Hence the chain rule $\frac{d z}{d x} = \frac{鈭� z}{鈭� x} + \frac{鈭� z}{鈭� y} \frac{d y}{d x}$ is proved.

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Most popular questions from this chapter

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