Chapter 2: Q. A (page 238)

Use the definition of the derivative to prove that the derivative of a sum of functions $f (x) + g (x)$ is equal to the sum of their derivatives $f' (x) + g' (x)$

Short Answer

Expert verified

$\frac{d}{d x} (f (x) + g (x)) = f' (x) + g' (x)$

Step by step solution

Step 1:Given information

The given expression is $f (x) + g (x)$

Step 2:Explanation

We’ll start with the sum of two functions. First recall the sum into the definition of the derivative and rewrite the numerator a little.

{(f (x) + g (x))}^{'} = \lim_{h \to 0} \frac{f (x + h) + g (x + h) - (f (x) + g (x))}{h}

$= \lim_{h \to 0} \frac{f (x + h) - f (x) + g (x + h) - g (x)}{h}$

Now, break up the fraction into two pieces and recall that the limit of a sum is the sum of the limits

\Rightarrow {(f (x) + g (x))}^{'} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} + \lim_{h \to 0} \frac{g (x + h) - g (x)}{h}

$= f' (x) + g' (x)$

The proof of the difference of two functions in nearly identical

{(f (x) - g (x))}^{'} = \lim_{h \to 0} \frac{f (x + h) - g (x + h) - (f (x) - g (x))}{h}

$= \lim_{h \to 0} \frac{f (x + h) - f (x) - g (x + h) - g (x)}{h}$

$\Rightarrow {(f (x) - g (x))}^{'} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} - \lim_{h \to 0} \frac{g (x + h) - g (x)}{h}$

$= f' (x) - g' (x)$

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Use the definition of the derivative to prove that the derivative of a sum of functions $f (x) + g (x)$ is equal to the sum of their derivatives $f' (x) + g' (x)$

Short Answer

Step by step solution

Step 1:Given information

Step 2:Explanation

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Use the definition of the derivative to prove that the derivative of a sum of functions fx+gxis equal to the sum of their derivativesf'x+g'x

Short Answer

Step by step solution

Step 1:Given information

Step 2:Explanation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

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Study anywhere. Anytime. Across all devices.

Use the definition of the derivative to prove that the derivative of a sum of functions $f (x) + g (x)$ is equal to the sum of their derivatives $f' (x) + g' (x)$