Find the Derivative - d/dk 7xe^(-kx)

$7 x e^{- k x}$

Step 1

Since $7 x$ is constant with respect to $k$ , the derivative of $7 x e^{- k x}$ with respect to $k$ is $7 x \frac{d}{d k} [e^{- k x}]$ .

$7 x \frac{d}{d k} [e^{- k x}]$

Step 2

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

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

$7 x (\frac{d}{d u} [e^{u}] \frac{d}{d k} [- k x])$

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

$7 x (e^{u} \frac{d}{d k} [- k x])$

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

$7 x (e^{- k x} \frac{d}{d k} [- k x])$

Step 3

Differentiate using the Constant Multiple Rule.

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

$7 x (e^{- k x} (- x \frac{d}{d k} [k]))$

Multiply $- 1$ by $7$ .

$- 7 x (e^{- k x} (x \frac{d}{d k} [k]))$

Step 4

Raise $x$ to the power of $1$ .

$- 7 (x^{1} x) (e^{- k x} (\frac{d}{d k} [k]))$

Step 5

Raise $x$ to the power of $1$ .

$- 7 (x^{1} x^{1}) (e^{- k x} (\frac{d}{d k} [k]))$

Step 6

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

$- 7 x^{1 + 1} (e^{- k x} (\frac{d}{d k} [k]))$

Step 7

Add $1$ and $1$ .

$- 7 x^{2} (e^{- k x} (\frac{d}{d k} [k]))$

Step 8

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

$- 7 x^{2} (e^{- k x} \cdot 1)$

Step 9

Multiply $e^{- k x}$ by $1$ .

$- 7 x^{2} e^{- k x}$