Answer:
![\displaystyle (d)/(dx)[e^(2x)] = 2e^(2x)](https://img.qammunity.org/2022/formulas/mathematics/college/1imqjn8t7osesvo9ghim8vmwymgs3jvgbt.png)
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Calculus
Derivatives
Derivative Notation
Derivative Rule [Chain Rule]:
![\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)](https://img.qammunity.org/2022/formulas/mathematics/high-school/vue68srn3fe6bds4idxorm97z7tgwelamw.png)
eˣ Derivative:
![\displaystyle (d)/(dx) [e^u]=e^u \cdot u'](https://img.qammunity.org/2022/formulas/mathematics/college/m3gc6cokukcgnox71iplq3x9xhifiqsdyo.png)
Explanation:
Step 1: Define
![\displaystyle (d)/(dx)[e^(2x)]](https://img.qammunity.org/2022/formulas/mathematics/college/xa8i0iv0pg8pvpikntu0bebfhe5wbajt4z.png)
Step 2: Differentiate
- [Derivative] eˣ Derivative [Chain Rule]:
![\displaystyle (d)/(dx)[e^(2x)] = 2x^(1 - 1) \cdot e^(2x)](https://img.qammunity.org/2022/formulas/mathematics/college/zdorc5xsqpmuruudzjltg608mj975mjuhw.png)
- [Derivative] Simplify:
![\displaystyle (d)/(dx)[e^(2x)] = 2x^(0) \cdot e^(2x)](https://img.qammunity.org/2022/formulas/mathematics/college/ogywpntqagsnjymxku1az3victfzof0jlp.png)
- [Derivative] Simplify:
![\displaystyle (d)/(dx)[e^(2x)] = 2(1) \cdot e^(2x)](https://img.qammunity.org/2022/formulas/mathematics/college/un5r6v2oh2vh2wkr3r5tyloc5mteocbdnp.png)
- [Derivative] Multiply:
![\displaystyle (d)/(dx)[e^(2x)] = 2 \cdot e^(2x)](https://img.qammunity.org/2022/formulas/mathematics/college/mjp547qsqw5gkp8skx6l8btx1mquecbkx3.png)
- [Derivative] Multiply:
![\displaystyle (d)/(dx)[e^(2x)] = 2e^(2x)](https://img.qammunity.org/2022/formulas/mathematics/college/1imqjn8t7osesvo9ghim8vmwymgs3jvgbt.png)
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Derivatives
Book: College Calculus 10e