Algebra Examples

$Q + (100 - Q) e^{- (k) (t)}$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $Q + (100 - Q) e^{- k t}$ with respect to $Q$ is $\frac{d}{d Q} [Q] + \frac{d}{d Q} [(100 - Q) e^{- k t}]$ .

$\frac{d}{d Q} [Q] + \frac{d}{d Q} [(100 - Q) e^{- k t}]$

Step 1.2

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

$1 + \frac{d}{d Q} [(100 - Q) e^{- k t}]$

Step 2

Evaluate $\frac{d}{d Q} [(100 - Q) e^{- k t}]$ .

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Step 2.1

Since $e^{- k t}$ is constant with respect to $Q$ , the derivative of $(100 - Q) e^{- k t}$ with respect to $Q$ is $e^{- k t} \frac{d}{d Q} [100 - Q]$ .

$1 + e^{- k t} \frac{d}{d Q} [100 - Q]$

Step 2.2

By the Sum Rule, the derivative of $100 - Q$ with respect to $Q$ is $\frac{d}{d Q} [100] + \frac{d}{d Q} [- Q]$ .

$1 + e^{- k t} (\frac{d}{d Q} [100] + \frac{d}{d Q} [- Q])$

Step 2.3

Since $100$ is constant with respect to $Q$ , the derivative of $100$ with respect to $Q$ is $0$ .

$1 + e^{- k t} (0 + \frac{d}{d Q} [- Q])$

Step 2.4

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

$1 + e^{- k t} (0 - \frac{d}{d Q} [Q])$

Step 2.5

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

$1 + e^{- k t} (0 - 1 \cdot 1)$

Step 2.6

Multiply $- 1$ by $1$ .

$1 + e^{- k t} (0 - 1)$

Step 2.7

Subtract $1$ from $0$ .

$1 + e^{- k t} \cdot - 1$

Step 2.8

Move $- 1$ to the left of $e^{- k t}$ .

$1 - 1 \cdot e^{- k t}$

Step 2.9

Rewrite $- 1 e^{- k t}$ as $- e^{- k t}$ .

$1 - e^{- k t}$