Calculus Examples

$10^{- x}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 10^{x}$ and $g (x) = - x$ .

Tap for more steps...

Step 1.1

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

$\frac{d}{d u} [10^{u}] \frac{d}{d x} [- x]$

Step 1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $10$ .

$10^{u} \ln (10) \frac{d}{d x} [- x]$

Step 1.3

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

$10^{- x} \ln (10) \frac{d}{d x} [- x]$

Step 2

Differentiate.

Tap for more steps...

Step 2.1

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

$10^{- x} \ln (10) (- \frac{d}{d x} [x])$

Step 2.2

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

$10^{- x} \ln (10) (- 1 \cdot 1)$

Step 2.3

Simplify the expression.

Tap for more steps...

Step 2.3.1

Multiply $- 1$ by $1$ .

$10^{- x} \ln (10) \cdot - 1$

Step 2.3.2

Move $- 1$ to the left of $10^{- x} \ln (10)$ .

$- 1 \cdot (10^{- x} \ln (10))$

Step 2.3.3

Rewrite $- 1 \cdot 10^{- x}$ as $- 10^{- x}$ .

$- 10^{- x} \ln (10)$