Problems with pi, arctan and such

[nitro_nay]

I'm doing a math practice test for grade 12 calculus right now (I have a test tomorrow) and one question has me stumped. Here's the question.

Let g be a function defined by g(x) = [Arctan(x)]/x
The value of g'(1) is [a + b(pi)]/4
Determine the value of a - b

I've tried several ways to solve this, but I've failed several times. Any idea on how to solve this equation (according to the teacher, the answer is 3)

 

[stapel]

Let g be a function defined by g(x) = [Arctan(x)]/x
The value of g'(1) is [a + b(pi)]/4
Determine the value of a - b

Find the derivative of g(x).

Evaluate g' at x = 1.

Determine the values of "a" and "b" (by comparison).

Evaluate "a - b".

I've tried several ways to solve this, but I've failed several times.

It's a shame you didn't show your work: there's no way for us to see where you might have gone wrong. Sorry!

Eliz.

 

[nitro_nay]

a brief history of the work I did goes something like this.

g'(x) = [x^2 - (x^2 + 1)(x)(arctanx)]/[(x^2+1)(x)(x^2)]
So g'(1) should be [1-2(pi/4)]/2

That means [1-2(pi/4)]/2 = [a + b(pi)]/4

I end up with 2 - 4(pi) = a + b(pi) and I dont know where to go from there

(by the way, Im not sure comparison would work. I only have one equation for the 2 variables)

 

[stapel]

a brief history of the work I did goes something like this.

g'(x) = [x^2 - (x^2 + 1)(x)(arctanx)]/[(x^2+1)(x)(x^2)]

I think the above means that you applied the Quotient Rule and, after some simplification (but not a complete simplification), got the result above. However, I cannot guess how you arrived at the final result above...?

Try writing the function as arctan(x) x^-1, and applying the Product Rule. Perhaps there will be fewer errors then...? And remember that the derivative of arctan(x) is 1/(1 + x²). :wink:

Eliz.

 

[nitro_nay]

I did

 

[nitro_nay]

never mind, a friend solved it for me. Thanks for the help though

 

[stapel]

I did

Um... "You did" what...? :shock:

A friend gave you the solution to this exercise, so you're done with this one. But, in future, please help us help you:

By communicating clearly and completely, we will have some idea what the actual question is, and what you've done with it. Having this information will greatly assist us in helping you next time. :wink:

Thank you!

Eliz.

 

[Deleted member 4993]

I'm doing a math practice test for grade 12 calculus right now (I have a test tomorrow) and one question has me stumped. Here's the question.

Let g be a function defined by g(x) = [Arctan(x)]/x
The value of g'(1) is [a + b(pi)]/4
Determine the value of a - b

I've tried several ways to solve this, but I've failed several times. Any idea on how to solve this equation (according to the teacher, the answer is 3)

\(\displaystyle g(x) = \frac {tan^{-1}(x)}{x}\)

\(\displaystyle g'(x) = \frac {{\frac{1}{1+x^2}\cdot x - tan^{-1}(x)}}{x^2}\)

\(\displaystyle g'(1) = \frac {{\frac{1}{1+1^2}\cdot 1 - tan^{-1}(1)}}{1^2}\)

g'(1) = 1/2 - π/4

so

a = 2

and

b = -1

and

a - b = 3