Calculus Examples

$\frac{d}{d x} \int_{0}^{x^{2}} \frac{1 - t}{1 - t^{2}} d t$

Step 1

Simplify the denominator.

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Step 1.1

Rewrite $1$ as $1^{2}$ .

$\frac{d}{d x} [\frac{d}{d x} \cdot \int_{0}^{x^{2}} \frac{1 - t}{1^{2} - t^{2}} d t]$

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = t$ .

$\frac{d}{d x} [\frac{d}{d x} \cdot \int_{0}^{x^{2}} \frac{1 - t}{(1 + t) (1 - t)} d t]$

Step 2

Reduce the expression by cancelling the common factors.

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Step 2.1

Cancel the common factor of $1 - t$ .

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Step 2.1.1

Cancel the common factor.

$\frac{d}{d x} [\frac{d}{d x} \cdot \int_{0}^{x^{2}} \frac{1 - t}{(1 + t) (1 - t)} d t]$

Step 2.1.2

Rewrite the expression.

$\frac{d}{d x} [\frac{d}{d x} \cdot \int_{0}^{x^{2}} \frac{1}{1 + t} d t]$

Step 2.2

Cancel the common factor of $d$ .

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Step 2.2.1

Cancel the common factor.

$\frac{d}{d x} [\frac{d}{d x} \cdot \int_{0}^{x^{2}} \frac{1}{1 + t} d t]$

Step 2.2.2

Rewrite the expression.

$\frac{d}{d x} [\frac{1}{x} \cdot \int_{0}^{x^{2}} \frac{1}{1 + t} d t]$

Step 3

Since $\int_{0}^{x^{2}} \frac{1}{1 + t} d t$ is constant with respect to $x$ , the derivative of $\frac{1}{x} \int_{0}^{x^{2}} \frac{1}{1 + t} d t$ with respect to $x$ is $\int_{0}^{x^{2}} \frac{1}{1 + t} d t \frac{d}{d x} [\frac{1}{x}]$ .

$\int_{0}^{x^{2}} \frac{1}{1 + t} d t \frac{d}{d x} [\frac{1}{x}]$

Step 4

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$\int_{0}^{x^{2}} \frac{1}{1 + t} d t \frac{d}{d x} [x^{- 1}]$

Step 5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 1$ .

$\int_{0}^{x^{2}} \frac{1}{1 + t} d t (- x^{- 2})$

Step 6

Simplify.

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Step 6.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\int_{0}^{x^{2}} \frac{1}{1 + t} d t (- \frac{1}{x^{2}})$

Step 6.2

Reorder the factors of $\int_{0}^{x^{2}} \frac{1}{1 + t} d t (- \frac{1}{x^{2}})$ .

$- \frac{1}{x^{2}} \int_{0}^{x^{2}} \frac{1}{1 + t} d t$

Step 6.3

Combine $- \frac{1}{x^{2}}$ and $\int_{0}^{x^{2}} \frac{1}{1 + t} d t$ .

$- \frac{\int_{0}^{x^{2}} \frac{1}{1 + t} d t}{x^{2}}$