Chapter 2: Q 35 (page 238)

Find a function $f$ that has the given derivative $f^{'}$ and value f(c).
$f' (x) = \frac{x}{1 + 4 x^{2}}, f (0) = 1$

Short Answer

Expert verified

The anti derivative is $f (x) = \frac{1}{8} \ln (1 + 4 x^{2}) + 1$

Step by step solution

Step 1:Given information

The given expression is $f' (x) = \frac{x}{1 + 4 x^{2}}$

Step 2:Simplification

Given

$f' (x) = \frac{x}{1 + 4 x^{2}}$

We will use an intelligent guess-and-check method to find f

Since differentiating a power function one and derivative of $\ln x i s \frac{1}{x}$

One can start with $f (x) = \ln (1 + 4 x^{2})$

Now,

$f' (x) = \frac{d}{d x} \{\ln (1 + 4 x^{2})\}$

$= \frac{1}{1 + 4 x^{2}} \frac{d}{d x} (1 + 4 x^{2})$

$= \frac{8 x}{1 + 4 x^{2}}$

Hence, we can guess the function as

$f (x) = \frac{1}{8} \ln (1 + 4 x^{2}) + c$

Now ,

f (0) = 1 \Rightarrow 1 = \ln (1 + 0) + c

$\Rightarrow c = 1$

Therefore,

$f (x) = \frac{1}{8} \ln (1 + 4 x^{2}) + 1$

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Find a function $f$ that has the given derivative $f^{'}$ and value f(c).
$f' (x) = \frac{x}{1 + 4 x^{2}}, f (0) = 1$

Short Answer

Step by step solution

Step 1:Given information

Step 2:Simplification

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Step by step solution

Step 1:Given information

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One App. One Place for Learning.

Most popular questions from this chapter

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Find a function $f$ that has the given derivative $f^{'}$ and value f(c).
$f' (x) = \frac{x}{1 + 4 x^{2}}, f (0) = 1$