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Chapter 2: Q 1. (page 237)

Find the derivatives of $f (x) = x^{2} .$

Short Answer

Expert verified

The derivative of $f (x) = x^{2}$ is $2 x .$

Step by step solution

Given Information

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

Simplification

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{{(x + h)}^{2} - x^{2}}{h} = \lim_{h \to 0} \frac{x^{2} + 2 h x + h^{2} - x^{2}}{h} = \lim_{h \to 0} \frac{(2 x + h) h}{h} = 2 x T h e r e f o r e, t h e d e r i v a t e o f f (x) = x^{2} i s 2 x .$

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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.

27. $f (x) = 1 - x^{3}, x = - 1$

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 the definition of the derivative to find $f$ for each function $f$ in Exercises 39-54

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

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 $x = - 3$

On earth, A falling object has a downward acceleration of 32 feet per second per second due to gravity. Suppose an object falls from an initial height of $s_{0}$ ,With an initial velocity of $v_{0}$ feet per second, Use antiderivatives to show that the equations for the position and velocity of the object after t seconds are respectively $s (t) = - 16 t^{2} + v_{0} t + s^{0}$ and $v (t) = - 32 t + v_{o} (t)$

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Find the derivatives off(x)=x2.

Short Answer

Step by step solution

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Find the derivatives of $f (x) = x^{2} .$