Creating A Derivative Problem that has a specific solution

In summary, the conversation discusses finding an equation using the quotient or power rule that will result in a derivative of either sec(x) or cot(x). The person providing help suggests using the chain rule and replacing u with cos(x) to solve this problem. The conversation also touches on the use of integrals and the derivatives of basic trig functions. The chain rule is explained and the person seeking help expresses understanding and gratitude for the explanation.
  • #1
Papa Hyman
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OP has received a warning about posting a question with no effort shown

Homework Statement


So this is a problem that I am at a complete loss with. The question asked is, give an equation, using the quotient or power rule that derivative is equal to either sec(x) or cot(x). It doesn't matter which one, sec(x) or cot(x), just as long as the initial equation's derivative is one of them

Thank you for the help
Papa
 
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  • #2
My first reaction was to use the integral or "anti-derivative" but I suspect this is an exercise preliminary to introducing the anti-derivative.

You should know the chain rule for differentiation and that the derivative of cos(x) is -sin(x) so you want to find a function, f(u), so that f'(u)= 1/u and then replace u with cos(x).
 
  • #3
Thank you for the quick reply!

That may be the case, but up until this point and in none of the notes provided for this unit has taught me even the slightest about integrals or the chain rule, could you possibly explain?

Thanks!
 
  • #4
So, you have learned the quotient rule and the product rule. And I assume you have learned the derivatives of the basic trig functions.
Your goal is to have a derivative that looks like ##\frac 1 {\cos x} ## or ##\frac {\cos x }{\sin x} ##.
Have you already learned about the derivative of ##\ln x ##?
The chain rule says if you have functions of functions i.e. ##f(g(x))## then ##\frac d{dx}f(g(x))=f'(g(x))g'(x)##.
 
  • #5
oooh, alright the chain rule makes sense like that. Thank you very much!
 

FAQ: Creating A Derivative Problem that has a specific solution

How do you create a derivative problem?

To create a derivative problem, you start by identifying a specific function or equation that has a known solution. Then, you use the rules of differentiation to find the derivative of that function or equation. This derivative can then be used to create a new problem that has the same solution as the original function.

What is the purpose of creating a derivative problem with a specific solution?

The purpose of creating a derivative problem with a specific solution is to test understanding and application of the rules of differentiation. It allows for practice in finding derivatives and applying them to new problems.

Can a derivative problem have more than one solution?

No, a derivative problem will have the same solution as the original function. However, there may be multiple ways to arrive at the same solution through different methods of differentiation.

How do you know if a derivative problem has a specific solution?

A derivative problem will have a specific solution if the original function has a known solution. This can be determined by checking if the function is a well-known type, such as a polynomial, trigonometric, or exponential function.

Are there any limitations to creating a derivative problem with a specific solution?

One limitation is that the original function must have a known solution in order to create a derivative problem with a specific solution. Additionally, the derivative problem may become more complex and difficult to solve depending on the complexity of the original function.

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