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Chapter 2: Q 10. (page 237)

Find the derivatives of the function: $f (x) = x^{- 1} {(4 - x)}^{2} .$

Short Answer

Expert verified

The derivative of $f (x) = x^{- 1} {(4 - x)}^{2} i s 1 - {(\frac{4}{x})}^{2} .$

Step by step solution

01

Given information

The given expression is $f (x) = x^{- 1} {(4 - x)}^{2} .$

02

Simplification

$f' (x) = \frac{d}{d x} {f (x)} = \frac{d}{d x} {x^{- 1} {(4 - x)}^{2}} = - \frac{1}{x^{2}} {(4 - x)}^{2} + \frac{1}{x} . 2 (4 - x) . - 1 = - {\frac{16 - 8 x + x^{2} + 2 x (4 - x)}{x^{2}}} = - (\frac{16 - 8 x + x^{2} + 8 x - 2 x^{2}}{x^{2}}) = - \frac{16 - x^{2}}{x^{2}} = 1 - {(\frac{4}{x})}^{2} H e n c e, f' (x) = 1 - {(\frac{4}{x})}^{2} .$

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Most popular questions from this chapter

Use (a) the $h \to 0$ definition of the derivative and then

(b) the $z \to c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23–38.

28. $f (x) = x^{4} + 1, x = 2$

Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating.

$f (x) = \frac{\frac{1}{x} - 3 x^{2}}{x^{5} - \frac{1}{\sqrt{x}}}$

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.

The tangent line to $f (x) = x^{2}$ at $x = - 3$

Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating.

$f (x) = x^{- \frac{1}{2}} {(x^{2} - 1)}^{3}$

Use (a) the $h \to 0$ definition of the derivative and then

(b) the $z \to c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23–38.

23. $f (x) = x^{2}, x = - 3$

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