Calculus Examples

$y = x^{2} \tanh (2 x - 1)$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = \tanh (2 x - 1)$ .

$x^{2} \frac{d}{d x} [\tanh (2 x - 1)] + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tanh (x)$ and $g (x) = 2 x - 1$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $2 x - 1$ .

$x^{2} (\frac{d}{d u} [\tanh (u)] \frac{d}{d x} [2 x - 1]) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 2.2

The derivative of $\tanh (u)$ with respect to $u$ is ${sech}^{2} (u)$ .

$x^{2} ({sech}^{2} (u) \frac{d}{d x} [2 x - 1]) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 2.3

Replace all occurrences of $u$ with $2 x - 1$ .

$x^{2} ({sech}^{2} (2 x - 1) \frac{d}{d x} [2 x - 1]) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$x^{2} ({sech}^{2} (2 x - 1) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1])) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 3.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$x^{2} ({sech}^{2} (2 x - 1) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1])) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 3.3

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

$x^{2} ({sech}^{2} (2 x - 1) (2 \cdot 1 + \frac{d}{d x} [- 1])) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 3.4

Multiply $2$ by $1$ .

$x^{2} ({sech}^{2} (2 x - 1) (2 + \frac{d}{d x} [- 1])) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 3.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$x^{2} ({sech}^{2} (2 x - 1) (2 + 0)) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 3.6

Simplify the expression.

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Step 3.6.1

Add $2$ and $0$ .

$x^{2} ({sech}^{2} (2 x - 1) \cdot 2) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 3.6.2

Move $2$ to the left of ${sech}^{2} (2 x - 1)$ .

$x^{2} (2 \cdot {sech}^{2} (2 x - 1)) + \tanh (2 x - 1) \frac{d}{d x} [x^{2}]$

Step 3.7

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$x^{2} (2 {sech}^{2} (2 x - 1)) + \tanh (2 x - 1) (2 x)$

Step 3.8

Reorder terms.

$2 x^{2} {sech}^{2} (2 x - 1) + 2 x \tanh (2 x - 1)$