# What is the first derivative of v(t)=(1+3^t)/3^t?

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We have to find the derivative of v(t)=(1+3^t)/3^t

Here we use the quotient rule, if f(x) = g(x) / h(x)

=> f'(x) = [g'(x)*h(x) - g(x)*h'(x)]/ (h(x))^2

So we have

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

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

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

=> v'(t) = ln 3( 1 - ( 1+ 3^t)/ 3^t)

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

=> v'(t) = -ln 3/ 3^t

**Therefore the required derivative is -ln 3/ 3^t**

Given the function:

v(t) = (1+3^t) /3^t

We need to find the first derivative v'(t).

First we will rewrite the fraction:

V(t) = 1/3^t + 3^t/3^t

= 3^-t + 1

==> v(t) =(1/3^t) +1

Now we will differentiate.

==> v'(t) = (1/3^t)' + (1)'

= (3^-t)' + 0

We will assume that u= -t ==> u' = - 1

==> v'(t) = (3^u)'

==> v'(t) = 3^u * ln3 * u'

= 3^-t * ln3 * -1

= -ln3 / 3^t

But we know that -ln3 = ln3^-1 = ln(1/3)

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

To find the first derivative of v(t)=(1+3^t)/3^t.

v(t) = (1+3^t)/3^t.

v(t) = 1/3^t + 1.

v'(t) = (3^-t +1)'

v'(t) = (3^(-t)+(1)'

v'(t) = (3^(-t)}(ln3) (-t)'+ 0, as (a^x) = (a^x)(lna)

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

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)**