BSME
1 Answer
melandrupert
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)
| 14 years ago. Rating: 1 | |
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