Equation of a tangent line given x

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Discussion Overview

The discussion revolves around finding the equation of the tangent line to the function \( y = \sqrt{x^4 - 2337} \) at the point where \( x = 7 \). Participants explore the necessary steps to derive the tangent line, including calculating the slope and the function value at the given point.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant seeks assistance in finding the tangent line equation given only an x value.
  • Another participant asks if the original poster knows how to find the slope of the tangent line.
  • A participant mentions the slope formula but notes the challenge of having only one point.
  • There is a discussion about the derivative of the function and the application of the chain rule.
  • One participant expresses uncertainty about the derivative calculation and seeks clarification on the correct approach.
  • Another participant suggests revising the chain rule and provides a hint about the derivative of the square root function.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the derivative calculation, and there is uncertainty regarding the correct application of the chain rule. Multiple viewpoints on how to approach the problem remain present.

Contextual Notes

Participants express varying levels of familiarity with calculus concepts, particularly the derivative and the chain rule, which may affect their understanding of the problem.

Drewnlauren06
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I am having a time trying to figure out this problem given only an x value. The equation is "Find the equation of the tangent line to y= Square root of x^4-2337 ...at x=7. If anybody could help me solve this it would be a great help. Thanks
 
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Welcome to PF!

Drewnlauren06 said:
"Find the equation of the tangent line to y= Square root of x^4-2337 ...at x=7.

Hi Drewnlauren06! Welcome to PF! :smile:

Do you know how to find the slope (the y/x) of the tangent line?​
 
I know to find Y you set x=0 I believe and I know the slope formula is y2-y1/x2-x1. I don't exactly remember how to pull everything together.
 
Drewnlauren06 said:
I know to find Y you set x=0 I believe and I know the slope formula is y2-y1/x2-x1. I don't exactly remember how to pull everything together.

yes … y2-y1/x2-x1 is the slope of a line, if you already know two points that it goes through …

but you only know one point …

have you done calculus? do you know what the derivative of √(x^4-2337) is? :smile:​
 
I'm in Business calculus now. If I was going to take the derivative of √(x^4-2337). I would get rid of the square root by (x^4-2337)^1/2 and the derivative would be 1/2(x^4-2337)^-1/2 or would it be [2x^3-(2337/2)]^-1/2 or neither I don't know. I know how to do basic derivatives with no problem but this one is just giving me problems.
 
Hi Drewnlauren06! :smile:

(have a square-root: √ and use the X2 tag, just above the reply box :wink:)
Drewnlauren06 said:
I'm in Business calculus now. If I was going to take the derivative of √(x^4-2337). I would get rid of the square root by (x^4-2337)^1/2

That's right :smile:
… and the derivative would be 1/2(x^4-2337)^-1/2 or would it be [2x^3-(2337/2)]^-1/2 or neither I don't know. I know how to do basic derivatives with no problem but this one is just giving me problems.

ah, you need to revise the chain rule …

the derivative of √(f(x)) is f'(x) times 1/2√(f(x)) …

so use that first formula of yours, but multiply it by the derivative of x4 - 2337 :wink:
 

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