Find derivative with exponential function?

In summary, the answer to the question is that $(f*g)'=f'*g' and the answer to the question of finding g'(x) if g(x)=sqrtx(ex) is .5x-1/2ex.
  • #1
coolbeans33
23
0
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?
 
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  • #2
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:
  • #3
Re: find derivative with exponential function?

ZaidAlyafey said:
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).
 
  • #4
Re: find derivative with exponential function?

coolbeans33 said:
...
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.
 
  • #5
Re: find derivative with exponential function?

MarkFL said:
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?
 
  • #6
Re: find derivative with exponential function?

coolbeans33 said:
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:
$$(fgh)'=f'gh+fg'h+fgh'.$$
 

FAQ: Find derivative with exponential function?

What is the exponential function?

The exponential function is a mathematical function in which the independent variable appears as an exponent. It is written in the form of f(x) = ax, where a is a constant base and x is the independent variable.

How do you find the derivative of an exponential function?

To find the derivative of an exponential function, you can use the rule d/dx(ax) = axln(a), where a is the constant base. This means that the derivative of an exponential function is equal to the original function multiplied by the natural logarithm of the base.

What is the purpose of finding the derivative of an exponential function?

The derivative of an exponential function helps us find the slope of the function at any point on its graph. This is useful in many applications, such as in finance and physics, where exponential functions are commonly used to model growth and decay.

Are there any special cases when finding the derivative of an exponential function?

Yes, there are two special cases when finding the derivative of an exponential function. The first is when the base a is equal to the number e (approximately 2.718), in which case the derivative simplifies to d/dx(ex) = ex. The second case is when the exponent is a constant, in which case the derivative becomes d/dx(ac) = cac-1, where c is the constant.

How is the derivative of an exponential function used in real life?

The derivative of an exponential function has many real-life applications. For example, it is used in finance to calculate compound interest and in physics to model radioactive decay. It is also used in biology to model population growth and in medicine to model drug dosage and absorption rates.

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