Calculating g`(π/3): An Attempted Solution

  • Thread starter SwedishFred
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In summary, the conversation is about a math problem where the goal is to find the derivative of a given function at a specific point. The attempted solution involved using the chain rule, but the final answer did not seem correct. The expert advises reviewing the chain rule and revisiting the problem.
  • #1
SwedishFred
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Homework Statement



Deside g`(π/3) when g(u)= (e^cosu-1/2)-6sin2u

Homework Equations


g`(π/3) when g(u)= (e^cosu-1/2)-6sin2u

3. The attempended solutions
ended upp with g`(u) = -12cosu-(e^cosu) sinu

g´(π/3)= -12cos(π/3)-(e^cos(π/3)) sin(π/3)
= -7.43 that dosen´t seem right..
 
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  • #2
SwedishFred said:

Homework Statement



Deside g`(π/3) when g(u)= (e^cosu-1/2)-6sin2u

Homework Equations


g`(π/3) when g(u)= (e^cosu-1/2)-6sin2u

3. The attempended solutions
ended upp with g`(u) = -12cosu-(e^cosu) sinu

g´(π/3)= -12cos(π/3)-(e^cos(π/3)) sin(π/3)
= -7.43 that dosen´t seem right..

You will need to use enough parentheses to remove ambiguity regarding your function.

Is the 1/2 an exponent?

Is that sin(2u) or is it sin2(u) ?

etc.
...

.
 
  • #3
Oki
g(u)=ecos(u)-1/2-6sin(2u)

g´(π/3)= -12cos(π/3)-ecos(π/3)sin(π/3)
= -7.43 that dosen´t seem right..
 
  • #4
SwedishFred said:
Oki
g(u)=ecos(u)-1/2-6sin(2u)

g´(π/3)= -12cos(π/3)-ecos(π/3)sin(π/3)
= -7.43 that dosen´t seem right..
It's not right.
 
  • #5
So i figure, where did i mess up , all is it all wrong ??
 
  • #6
SwedishFred said:
So i figure, where did i mess up , all is it all wrong ??

You made errors when using the chain rule (twice). I would go back and review that and revisit this problem after.
 

FAQ: Calculating g`(π/3): An Attempted Solution

1. How do you calculate g`(π/3)?

To calculate g`(π/3), you need to use the formula g`(x) = lim(h->0) [g(x+h) - g(x)] / h. In this case, plug in π/3 for x and simplify the equation to find the derivative.

2. What is the significance of calculating g`(π/3)?

Calculating g`(π/3) allows us to find the slope of the tangent line at x=π/3 on the graph of g(x). This can help us understand the behavior of the function and make predictions about its values near x=π/3.

3. Can you provide an example of calculating g`(π/3)?

Sure, let's say g(x) = x^2 + 3x. To find g`(π/3), we use the formula g`(x) = lim(h->0) [g(x+h) - g(x)] / h. Plugging in π/3 for x, we get g`(π/3) = lim(h->0) [(π/3 + h)^2 + 3(π/3 + h) - (π^2/9 + 3π/3)] / h. Simplifying this equation gives us g`(π/3) = 2π/3 + 3.

4. What is the domain of g`(x)?

The domain of g`(x) is the same as the domain of g(x). In this case, since we are calculating g`(π/3), the domain would be all real numbers except for x=π/3, as this would result in a division by zero in the formula.

5. How can we use g`(π/3) to approximate the value of g(π/3)?

We can use the equation g(x+h) = g(x) + g`(x)*h to find an approximation of g(π/3). Plugging in π/3 for x and a small value for h, we can use g`(π/3) to find the value of g(π/3 + h). Repeating this process with smaller and smaller values of h can help us get closer and closer to the actual value of g(π/3).

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