Integral and Derivative Problems

In summary, The conversation is about two math problems that the students are struggling with. The first problem involves integrating a complicated expression, but the solution is to split it into two separate functions. The second problem involves finding the derivative of a function, and the students are having trouble simplifying it. They receive advice on how to factor and simplify the expression, and eventually call their teacher for help.
  • #1
rwisz
2,268
0

Homework Statement


2 Questions here. Please help, math test tomorrow, study group failed for these 2 problems.
[tex]\int{\frac{1-sin\theta}{cos\theta}d\theta}[/tex]

and

Find [tex]\frac{dy}{dx} [/tex] if [tex]y=x^ee^x[/tex]

Homework Equations


NONE

The Attempt at a Solution


We've attempted every "u-substitution" we can think of for problem 1.

We've tried to differentiate problem 2 many times, and can't come up with the proper derivative.

thanks! Any advice is greatly appreciated.
 
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  • #2
My recomendation for the first one is to split the integral into two separate functions. Then all you have to integrate is sec(x) - tan(x) which is quite simple.

For the second problem show what work you've done as it should just be an application of the product rule.
 
  • #3
Thanks for the advice on Number 1! We integrated successfully!

On to number 2 however, here's what we have so far:

Using simply the product rule:
[tex]y'=ex^{e-1}e^x+x^ee^x[/tex]

We've attempted to simplify down etc, but not coming up with the right answer here!
Is it perhaps using the idea a base besides e?

We're really struggling here, this is our last problem!
 
  • #4
Why can't you find the right answer? Factor out e^x*x^(e-1) from both parts
 
  • #5
We're both looking at your reply with the most blank stare you can ever imagine...the [tex]x^{e-1}[/tex] is not in the second term whatsoever... Am I really as stupid as you're making me feel? I'm very confused here...
 
  • #6
As far as I can tell, the solution should be correct - aside from a restricted domain. Some thoughts on simplifications though: Factor (e^x)(x^e) out so you are left with [(e^x)(x^e)](e/x +1). That's about as simple as you can get it.
 
  • #7
x^e = x^(e-1) * x
 
  • #8
jgens said:
As far as I can tell, the solution should be correct - aside from a restricted domain. Some thoughts on simplifications though: Factor (e^x)(x^e) out so you are left with [(e^x)(x^e)](e/x +1). That's about as simple as you can get it.

I would factor out the smallest of both powers which in this case is e-1 for the power of x to get

[tex] ex^{e-1}e^x+x^ee^x = e^{x}x^{e-1}\left(e + x\right) [/tex]
 
  • #9
We opted on calling our teacher as well. And she correctly guided us through the problem... and somehow it worked. I doubt she'll give us one like this on the test... but thanks for all your help, much obliged.

Thanks!
 
  • #10
Either works, though I suppose your approach yields a simpler expression.
 
  • #11
rwisz said:
We're both looking at your reply with the most blank stare you can ever imagine...the [tex]x^{e-1}[/tex] is not in the second term whatsoever... Am I really as stupid as you're making me feel? I'm very confused here...

[itex]e^{x-1}[/itex] certainly is in the second term: the second term includes [itex]e^x[/itex] and [itex]e^x= (e^{x-1})(e)[/itex]
 
  • #12
You mean x^(e-1)?
 
  • #13
The first integral can even be done by simple substitution i.e
[tex]
\int{\frac{1-sin\theta}{cos\theta}d\theta}[/tex]
can be written as,
[tex]\int \frac{(sin {\frac {\theta}{2}}-~cos{\frac {\theta}{2}})^2}{(cos^2{\frac {\theta}{2}}-sin^2{\frac {\theta}{2}})}[/tex]
 
Last edited:

1. What is the difference between an integral and a derivative?

An integral is the inverse operation of a derivative, representing the accumulation of a function over a given interval. A derivative, on the other hand, represents the instantaneous rate of change of a function at a specific point.

2. How are integrals and derivatives used in real-world applications?

Integrals and derivatives are commonly used in physics, engineering, and economics to model and solve real-world problems. For example, derivatives are used to calculate the velocity and acceleration of an object, while integrals are used to find the total distance traveled by that object.

3. What are some common techniques for solving integral and derivative problems?

Some common techniques for solving integral and derivative problems include the power rule, substitution, integration by parts, and partial fractions. These techniques can be used to simplify complex functions and make them easier to integrate or differentiate.

4. Can integrals and derivatives be used to find the area and volume of a shape?

Yes, integrals can be used to find the area under a curve and the volume of a solid by using the Fundamental Theorem of Calculus. This theorem states that the integral of a function can be calculated by finding the antiderivative of that function, making it a useful tool for finding areas and volumes.

5. How do you determine which technique to use when solving an integral or derivative problem?

The best technique to use when solving an integral or derivative problem depends on the complexity of the function and the available techniques. It is important to have a good understanding of the different techniques and their strengths and weaknesses in order to choose the most efficient method for a specific problem.

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