Calculus Examples

$f (x) = 3 \sin (\frac{1 - x}{π})$

Find the first derivative.

Tap for more steps...

Since $3$ is constant with respect to $x$ , the derivative of $3 \sin (\frac{1 - x}{π})$ with respect to $x$ is $3 \frac{d}{d x} [\sin (\frac{1 - x}{π})]$ .

$f' (x) = 3 \frac{d}{d x} (\sin (\frac{1 - x}{π}))$

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

Tap for more steps...

To apply the Chain Rule, set $u$ as $\frac{1 - x}{π}$ .

$f' (x) = 3 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (\frac{1 - x}{π}))$

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$f' (x) = 3 (\cos (u) \frac{d}{d x} (\frac{1 - x}{π}))$

Replace all occurrences of $u$ with $\frac{1 - x}{π}$ .

$f' (x) = 3 (\cos (\frac{1 - x}{π}) \frac{d}{d x} (\frac{1 - x}{π}))$

Differentiate.

Tap for more steps...

Since $\frac{1}{π}$ is constant with respect to $x$ , the derivative of $\frac{1 - x}{π}$ with respect to $x$ is $\frac{1}{π} \frac{d}{d x} [1 - x]$ .

$f' (x) = 3 \cos (\frac{1 - x}{π}) (\frac{1}{π} \cdot \frac{d}{d x} (1 - x))$

Combine fractions.

Tap for more steps...

Combine $\frac{1}{π}$ and $3$ .

$f' (x) = \frac{3}{π} \cdot \cos (\frac{1 - x}{π}) \frac{d}{d x} (1 - x)$

Combine $\frac{3}{π}$ and $\cos (\frac{1 - x}{π})$ .

$f' (x) = \frac{3 \cos (\frac{1 - x}{π})}{π} \cdot \frac{d}{d x} (1 - x)$

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

$f' (x) = \frac{3 \cos (\frac{1 - x}{π})}{π} \cdot (\frac{d}{d x} (1) + \frac{d}{d x} (- x))$

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

$f' (x) = \frac{3 \cos (\frac{1 - x}{π})}{π} \cdot (0 + \frac{d}{d x} (- x))$

Add $0$ and $\frac{d}{d x} [- x]$ .

$f' (x) = \frac{3 \cos (\frac{1 - x}{π})}{π} \cdot \frac{d}{d x} (- x)$

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$f' (x) = \frac{3 \cos (\frac{1 - x}{π})}{π} \cdot (- \frac{d}{d x} x)$

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

$f' (x) = \frac{3 \cos (\frac{1 - x}{π})}{π} \cdot (- 1 \cdot 1)$

Combine fractions.

Tap for more steps...

Multiply $- 1$ by $1$ .

$f' (x) = \frac{3 \cos (\frac{1 - x}{π})}{π} \cdot - 1$

Combine $\frac{3 \cos (\frac{1 - x}{π})}{π}$ and $- 1$ .

$f' (x) = \frac{3 \cos (\frac{1 - x}{π}) \cdot - 1}{π}$

Simplify the expression.

Tap for more steps...

Multiply $- 1$ by $3$ .

$f' (x) = \frac{- 3 \cos (\frac{1 - x}{π})}{π}$

Move the negative in front of the fraction.

$f' (x) = - \frac{3 \cos (\frac{1 - x}{π})}{π}$

Find the second derivative.

Tap for more steps...

Since $- \frac{3}{π}$ is constant with respect to $x$ , the derivative of $- \frac{3 \cos (\frac{1 - x}{π})}{π}$ with respect to $x$ is $- \frac{3}{π} \frac{d}{d x} [\cos (\frac{1 - x}{π})]$ .

$f'' (x) = - \frac{3}{π} \cdot \frac{d}{d x} \cos (\frac{1 - x}{π})$

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

Tap for more steps...

To apply the Chain Rule, set $u$ as $\frac{1 - x}{π}$ .

$f'' (x) = - \frac{3}{π} \cdot (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (\frac{1 - x}{π}))$

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$f'' (x) = - \frac{3}{π} \cdot (- \sin (u) \frac{d}{d x} \frac{1 - x}{π})$

Replace all occurrences of $u$ with $\frac{1 - x}{π}$ .

$f'' (x) = - \frac{3}{π} \cdot (- \sin (\frac{1 - x}{π}) \frac{d}{d x} \frac{1 - x}{π})$

Differentiate using the Constant Multiple Rule.

Tap for more steps...

Multiply $- 1$ by $- 1$ .

$f'' (x) = 1 (\frac{3}{π}) \cdot (\sin (\frac{1 - x}{π}) \frac{d}{d x} (\frac{1 - x}{π}))$

Combine fractions.

Tap for more steps...

Multiply $\frac{3}{π}$ by $1$ .

$f'' (x) = \frac{3}{π} \cdot (\sin (\frac{1 - x}{π}) \frac{d}{d x} (\frac{1 - x}{π}))$

Combine $\sin (\frac{1 - x}{π})$ and $\frac{3}{π}$ .

$f'' (x) = \frac{\sin (\frac{1 - x}{π}) \cdot 3}{π} \cdot \frac{d}{d x} (\frac{1 - x}{π})$

Move $3$ to the left of $\sin (\frac{1 - x}{π})$ .

$f'' (x) = \frac{3 \cdot \sin (\frac{1 - x}{π})}{π} \cdot \frac{d}{d x} (\frac{1 - x}{π})$

Since $\frac{1}{π}$ is constant with respect to $x$ , the derivative of $\frac{1 - x}{π}$ with respect to $x$ is $\frac{1}{π} \frac{d}{d x} [1 - x]$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π} \cdot (\frac{1}{π} \cdot \frac{d}{d x} (1 - x))$

Multiply $\frac{1}{π}$ by $\frac{3 \sin (\frac{1 - x}{π})}{π}$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π π} \cdot \frac{d}{d x} (1 - x)$

Raise $π$ to the power of $1$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π π} \cdot \frac{d}{d x} (1 - x)$

Raise $π$ to the power of $1$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π π} \cdot \frac{d}{d x} (1 - x)$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π^{1 + 1}} \cdot \frac{d}{d x} (1 - x)$

Add $1$ and $1$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π^{2}} \cdot \frac{d}{d x} (1 - x)$

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

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π^{2}} \cdot (\frac{d}{d x} (1) + \frac{d}{d x} (- x))$

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

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π^{2}} \cdot (0 + \frac{d}{d x} (- x))$

Add $0$ and $\frac{d}{d x} [- x]$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π^{2}} \cdot \frac{d}{d x} (- x)$

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π^{2}} \cdot (- \frac{d}{d x} x)$

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

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π^{2}} \cdot (- 1 \cdot 1)$

Combine fractions.

Tap for more steps...

Multiply $- 1$ by $1$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π})}{π^{2}} \cdot - 1$

Combine $\frac{3 \sin (\frac{1 - x}{π})}{π^{2}}$ and $- 1$ .

$f'' (x) = \frac{3 \sin (\frac{1 - x}{π}) \cdot - 1}{π^{2}}$

Simplify the expression.

Tap for more steps...

Multiply $- 1$ by $3$ .

$f'' (x) = \frac{- 3 \sin (\frac{1 - x}{π})}{π^{2}}$

Move the negative in front of the fraction.

$f'' (x) = - \frac{3 \sin (\frac{1 - x}{π})}{π^{2}}$

The second derivative of $f (x)$ with respect to $x$ is $- \frac{3 \sin (\frac{1 - x}{π})}{π^{2}}$ .

$- \frac{3 \sin (\frac{1 - x}{π})}{π^{2}}$