Calculus Examples

$f (x) = \sin (1 - x)$

Step 1

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 $g (x) = 1 - x$ .

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Step 1.1

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

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [1 - x]$

Step 1.2

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

$\cos (u) \frac{d}{d x} [1 - x]$

Step 1.3

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

$\cos (1 - x) \frac{d}{d x} [1 - x]$

Step 2

Differentiate.

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Step 2.1

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

$\cos (1 - x) (\frac{d}{d x} [1] + \frac{d}{d x} [- x])$

Step 2.2

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

$\cos (1 - x) (0 + \frac{d}{d x} [- x])$

Step 2.3

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

$\cos (1 - x) \frac{d}{d x} [- x]$

Step 2.4

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

$\cos (1 - x) (- \frac{d}{d x} [x])$

Step 2.5

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

$\cos (1 - x) (- 1 \cdot 1)$

Step 2.6

Simplify the expression.

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Step 2.6.1

Multiply $- 1$ by $1$ .

$\cos (1 - x) \cdot - 1$

Step 2.6.2

Move $- 1$ to the left of $\cos (1 - x)$ .

$- 1 \cdot \cos (1 - x)$

Step 2.6.3

Rewrite $- 1 \cos (1 - x)$ as $- \cos (1 - x)$ .

$- \cos (1 - x)$