f(x) = e^x / (x+2)

Let f(x) = u/v such that:

u= e^x ==> u'= e^x

v= (x+2) ==> v' = 1

Noe by definition:

f'(x) = (u'v-uv')/v^2

=[ (e^x)*(x+2) - (e^x)*1]/(x+2)^2

= (e^x)(x+2 -1) /(x+2)^2

= (e^x)(x+1)/(x+2)^2

f'(0) = (e^0)(0+1)/(0+2)^2

= 1*1/2^2 = 1/4

==> f'(0) = 1/4

We'll calculate the first derivative of f(x) using the quotient rule. We notice that the numerator is an elementary exponential function and the denominator is a linear function.

The rule of quotient is:

f'(x) = [e^x / (x+2)]'

f'(x) = [(e^x)'*(x+2) - (e^x)*(x+2)']/(x+2)^2

f'(x) = [(e^x)*(x+2) - (e^x)]/(x+2)^2

We'll factorize by e^x:

f'(x) = (e^x)*(x+2-1)/(x+2)^2

f'(x) = (e^x)*(x+1)/(x+2)^2

Now, we'll put x=0 in the expression of the first derivative:

f'(0) = (e^0)*(0+1)/(0+2)^2

f'(0) = 1*1/2^2

**f'(0) = 1/4**

f(x) = e^x/(x+2).

To find f'(x).

Solution:

We know that the diffrential coefficient o fe^x = (e^x)' = e^x itself.

The differential coefficient of 1/(x+2) = {1/(x+2)}' = -1/(x+2)^2.

Also {u(x)/v(x)}' = u'(x)/v(x) +u(x)*([/v(x)]' = [u'(x)*v(v) - u(x)*v'(x)]/[v(x)]^2

So {(e^x)/(x+2)}' = {(e^x)'(x+2) - e^x*(x+2),}/(x+2)^2

= {e^x(x+2) -e^x *(1)}/(x+2)^2 = {(x+2-1)e^x}/(x+2)^2

f'(x)={(x+1)e^x}/(x+2)^2

f'(0) = {(0+1)e^0}/(0+1)^2 = 1/1 = 1

f'(0) = 1.