Differentiating a exponential function.

**vivek_master146** · January 31st 2011, 02:54 AM

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

I tried to solve it as follows:-

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

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

Using chain rule my answer is:-

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

Is it correct.? The actual answer is different.

**FernandoRevilla** · January 31st 2011, 02:59 AM

Originally Posted by vivek_master146

Is it correct.? The actual answer is different.

You can simplify:

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

and your function is:

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

not

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

Fernando Revilla

**vivek_master146** · January 31st 2011, 03:17 AM

Originally Posted by FernandoRevilla

You can simplify:

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

and your function is:

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

not

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

Thanks for reply. The first line totally went over my head. How did you derive $x^2$ .

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

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

**FernandoRevilla** · January 31st 2011, 03:22 AM

Originally Posted by vivek_master146

Thanks for reply. The first line totally went over my head. How did you derive $x^2$ .

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

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

The second one.

Fernando Revilla

**HallsofIvy** · January 31st 2011, 06:11 AM

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

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

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