# Thread: Differentiating a exponential function.

1. ## Differentiating a exponential function.

Differentiate (w.r.t x ) $\displaystyle e^{2logx + 3x}$

I tried to solve it as follows:-

$\displaystyle e^{2logx} + e^{3x}$

$\displaystyle \frac{\mbox{d}}{\mbox{dx}}(e^{2logx}) + \frac{\mbox{d}}{\mbox{dx}}(e^{3x})$

Using chain rule my answer is:-

$\displaystyle e^{2logx}. \frac{\ 2}{x} + e^{3x}.3$

Is it correct.? The actual answer is different.

2. Originally Posted by vivek_master146
Is it correct.? The actual answer is different.

You can simplify:

$\displaystyle e^{2\log x}=e^{\log x^2}=x^2$ .

$\displaystyle f(x)=x^2e^{3x}$

not

$\displaystyle f(x)=x^2+e^{3x}$

Fernando Revilla

3. Originally Posted by FernandoRevilla
You can simplify:

$\displaystyle e^{2\log x}=e^{\log x^2}=x^2$ .

$\displaystyle f(x)=x^2e^{3x}$

not

$\displaystyle f(x)=x^2+e^{3x}$
Thanks for reply. The first line totally went over my head. How did you derive $\displaystyle x^2$.

One more thing:- $\displaystyle e^{a +b} = e^a + e^b \ OR \ e^a.e^b$

Sorry for simple questions. I am directly studying graduation level maths without studying higher secondary maths.

4. Originally Posted by vivek_master146
Thanks for reply. The first line totally went over my head. How did you derive $\displaystyle x^2$.

By definition of natural logarithm: $\displaystyle e^{\log A}=A$

One more thing:- $\displaystyle e^{a +b} = e^a + e^b \ OR \ e^a.e^b$

The second one.

Fernando Revilla

5. You really should learn the "laws of exponents" better before attempting calculus.

$\displaystyle e^{a+ b}= e^ae^b$, not $\displaystyle e^a+ e^b$,
$\displaystyle (e^a)^b= e^{ab}$,
and, since $\displaystyle e^x$ and ln(x) are "inverse" functions,
$\displaystyle e^{ln(x)}= x$