Answer :
Sure, let's differentiate the given expression step-by-step.
We want to find the derivative of the expression [tex]\( \frac{e^{ax}}{1 + ax} \)[/tex] with respect to [tex]\( x \)[/tex]. To do this, we can apply the quotient rule for differentiation, which states:
[tex]\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2} \][/tex]
where [tex]\( u(x) = e^{ax} \)[/tex] and [tex]\( v(x) = 1 + ax \)[/tex].
First, let's find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex].
1. Derivative of [tex]\( u(x) \)[/tex]:
[tex]\[ u(x) = e^{ax} \][/tex]
Using the chain rule:
[tex]\[ u'(x) = \frac{d}{dx} (e^{ax}) = ae^{ax} \][/tex]
2. Derivative of [tex]\( v(x) \)[/tex]:
[tex]\[ v(x) = 1 + ax \][/tex]
Since [tex]\( v(x) \)[/tex] is a linear function with respect to [tex]\( x \)[/tex]:
[tex]\[ v'(x) = \frac{d}{dx} (1 + ax) = a \][/tex]
Now, we can apply the quotient rule:
[tex]\[ \frac{d}{dx} \left( \frac{e^{ax}}{1 + ax} \right) = \frac{(1 + ax) \cdot ae^{ax} - e^{ax} \cdot a}{(1 + ax)^2} \][/tex]
Let's simplify the numerator:
[tex]\[ (1 + ax) \cdot ae^{ax} - e^{ax} \cdot a = ae^{ax} (1 + ax) - ae^{ax} = ae^{ax} + a^2 xe^{ax} - ae^{ax} \][/tex]
[tex]\[ = a^2 xe^{ax} \][/tex]
So, the derivative is:
[tex]\[ \frac{d}{dx} \left( \frac{e^{ax}}{1 + ax} \right) = \frac{a e^{ax} + a^2 x e^{ax} - a e^{ax}}{(1 + ax)^2} = \frac{a^2 xe^{ax}}{(1 + ax)^2} \][/tex]
To summarize:
[tex]\[ \frac{d}{dx} \left( \frac{e^{ax}}{1 + ax} \right) = \frac{a e^{ax}}{1 + ax} - \frac{a e^{ax}}{(1 + ax)^2} \][/tex]
Or more neatly rewritten:
[tex]\[ \frac{d}{dx} \left( \frac{e^{ax}}{1 + ax} \right) = \frac{a e^{ax}}{1 + ax} - \frac{a e^{ax}}{(1 + ax)^2} \][/tex]
Therefore, the result is:
[tex]\[ \boxed{\frac{a e^{ax}}{1 + ax} - \frac{a e^{ax}}{(1 + ax)^2}} \][/tex]
We want to find the derivative of the expression [tex]\( \frac{e^{ax}}{1 + ax} \)[/tex] with respect to [tex]\( x \)[/tex]. To do this, we can apply the quotient rule for differentiation, which states:
[tex]\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2} \][/tex]
where [tex]\( u(x) = e^{ax} \)[/tex] and [tex]\( v(x) = 1 + ax \)[/tex].
First, let's find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex].
1. Derivative of [tex]\( u(x) \)[/tex]:
[tex]\[ u(x) = e^{ax} \][/tex]
Using the chain rule:
[tex]\[ u'(x) = \frac{d}{dx} (e^{ax}) = ae^{ax} \][/tex]
2. Derivative of [tex]\( v(x) \)[/tex]:
[tex]\[ v(x) = 1 + ax \][/tex]
Since [tex]\( v(x) \)[/tex] is a linear function with respect to [tex]\( x \)[/tex]:
[tex]\[ v'(x) = \frac{d}{dx} (1 + ax) = a \][/tex]
Now, we can apply the quotient rule:
[tex]\[ \frac{d}{dx} \left( \frac{e^{ax}}{1 + ax} \right) = \frac{(1 + ax) \cdot ae^{ax} - e^{ax} \cdot a}{(1 + ax)^2} \][/tex]
Let's simplify the numerator:
[tex]\[ (1 + ax) \cdot ae^{ax} - e^{ax} \cdot a = ae^{ax} (1 + ax) - ae^{ax} = ae^{ax} + a^2 xe^{ax} - ae^{ax} \][/tex]
[tex]\[ = a^2 xe^{ax} \][/tex]
So, the derivative is:
[tex]\[ \frac{d}{dx} \left( \frac{e^{ax}}{1 + ax} \right) = \frac{a e^{ax} + a^2 x e^{ax} - a e^{ax}}{(1 + ax)^2} = \frac{a^2 xe^{ax}}{(1 + ax)^2} \][/tex]
To summarize:
[tex]\[ \frac{d}{dx} \left( \frac{e^{ax}}{1 + ax} \right) = \frac{a e^{ax}}{1 + ax} - \frac{a e^{ax}}{(1 + ax)^2} \][/tex]
Or more neatly rewritten:
[tex]\[ \frac{d}{dx} \left( \frac{e^{ax}}{1 + ax} \right) = \frac{a e^{ax}}{1 + ax} - \frac{a e^{ax}}{(1 + ax)^2} \][/tex]
Therefore, the result is:
[tex]\[ \boxed{\frac{a e^{ax}}{1 + ax} - \frac{a e^{ax}}{(1 + ax)^2}} \][/tex]
