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Find derivative with exponential function?

coolbeans33

New member
Sep 19, 2013
23
f(x)=x2ex

the answer is f'(x)=(x2 + 2x)ex but I don't understand how to get there.

Also I need to find g'(x) if g(x)=sqrtx(ex)

would the answer for the second one be .5x-1/2ex?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: find derivative with exponential function?

Let \(\displaystyle f,g \) be defferentiable functions then

\(\displaystyle (f*g)'=f'*g'\)

EDIT : This is wrong , illustrated below .
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,905
Re: find derivative with exponential function?

Let \(\displaystyle f,g \) be defferentiable functions then

\(\displaystyle (f*g)'=f'*g'\)
Erm... I hope you meant to rectify that and say that $(f \cdot g)'=f' \cdot g + f \cdot g'$ (product rule).
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: find derivative with exponential function?

...
Also I need to find g'(x) if g(x)=sqrtx(ex)

would the answer for the second one be .5x-1/2ex?
No. Try applying the product rule for differentiation as stated by I like Serena. What do you find? Show your work, and if you have made a mistake, we will know where it is, and can then offer guidance to help correct the error in the application of the rule.
 

coolbeans33

New member
Sep 19, 2013
23
Re: find derivative with exponential function?

No. Try applying the product rule for differentiate as stated by I like Serena. What do you find? Show your work, and if you have made a mistake, we will know where it is, and can then offer guidance to help correct the error in the application of the rule.
ok so I used the product rule for this one, and I got (ex)(sqrt x) * (.5x1/2)(ex)

is this right?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: find derivative with exponential function?

ok so I used the product rule for this one, and I got (ex)(sqrt x) * (.5x1/2)(ex)

is this right?
It would be correct if your '*' changed to a '+', and your exponent on the second term was negative. That is,
$$(\sqrt{x} \, e^{x})'=\frac{1}{2\sqrt{x}} e^{x}+\sqrt{x} \, e^{x}=e^{x} \left( \frac{1}{2\sqrt{x}} +\sqrt{x} \right).$$

The way I think of the product rule is this: write down two copies of the product, add them together, and take a different derivative each time. This way of thinking about has the virtue of scalability:
$$(fgh)'=f'gh+fg'h+fgh'.$$