# 1 Answer

**We'll differentiate the given function, with respect to t.**

**We'll use the quotient rule:**

**v'(t) = [(1+3^t)'*(3^t) - (1+3^t)*(3^t)']/(3^t)^2**

**We'll differentiate and we'll get:**

**v'(t) = [(3^t*ln3)*(3^t) - (3^t*ln3)*(1+3^t)]/(3^t)^2**

**v'(t) = [(3^t*ln3)*(3^t - 1 -3^t)]/(3^t)^2**

**We'll eliminate like terms from numerator:**

**v'(t) = -(3^t*ln3)/(3^t)^2**

**We'll simplify and we'll get:**

**v'(t) = -(ln3)/(3^t)**

**v'(t) = (ln 1/3)/(3^t)**

**The first derivative of v(t)=(1+3^t)/3^t is:**

**v'(t) = (ln 1/3)/(3^t)**

12 years ago. Rating: 1 | |

### Top contributors in Mathematics category

### Unanswered Questions

new88100

Answers: 0
Views: 4
Rating: 0

tinhdau hoalai

Answers: 0
Views: 3
Rating: 0

Nha cai fb886site

Answers: 0
Views: 6
Rating: 0

hb88supply

Answers: 0
Views: 5
Rating: 0

bxbacvinh

Answers: 0
Views: 5
Rating: 0

VZ99

Answers: 0
Views: 8
Rating: 0

WW88 – Thiên đường cá cược hot nhất 2024

Answers: 0
Views: 8
Rating: 0

qh88f8com

> More questions...
Answers: 0
Views: 7
Rating: 0