Find derivative with exponential function?

[coolbeans33]

f(x)=x²e^x

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

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

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

 

[alyafey22]

Re: find derivative with exponential function?

Let $$f,g $$ be defferentiable functions then

$$(f*g)'=f'*g'$$

EDIT : This is wrong , illustrated below .

 

[I like Serena]

Re: find derivative with exponential function?

Let $$f,g $$ be defferentiable functions then

$$(f*g)'=f'*g'$$

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

 

[MarkFL]

Re: find derivative with exponential function?

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

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

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]

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 (e^x)(sqrt x) * (.5x^1/2)(e^x)

is this right?

 

[Ackbach]

Re: find derivative with exponential function?

ok so I used the product rule for this one, and I got (e^x)(sqrt x) * (.5x^1/2)(e^x)

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