# Find derivative with exponential function?

#### coolbeans33

##### New member
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
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 .

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
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

Staff member
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
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
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:
$$(fgh)'=f'gh+fg'h+fgh'.$$