Equation of a tangent line given x

In summary: The derivative of √(x^4-2337) is f'(x) times 1/2√(f(x)) … so use that first formula of yours, but multiply it by the derivative of x4 - 2337
  • #1
Drewnlauren06
4
0
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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  • #2
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?​
 
  • #3
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.
 
  • #4
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:​
 
  • #5
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.
 
  • #6
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:
 

FAQ: Equation of a tangent line given x

What is the equation of a tangent line given x?

The equation of a tangent line given x is y = f'(x)(x - x0) + f(x0), where f'(x) is the derivative of the function f(x) and x0 is the point at which the tangent line is drawn.

How do I find the slope of a tangent line given x?

The slope of a tangent line given x is equal to the derivative of the function at the given point. To find the derivative, you can use the power rule, product rule, quotient rule, or chain rule depending on the function.

What is the significance of the equation of a tangent line given x?

The equation of a tangent line given x allows us to find the instantaneous rate of change of a function at a specific point. This is useful in understanding the behavior of a function and making predictions about its values.

Can the equation of a tangent line given x be used to find the slope of a curve at any point?

Yes, the equation of a tangent line given x can be used to find the slope of a curve at any point, as long as the function is differentiable at that point.

Is there a specific method for finding the equation of a tangent line given x?

Yes, there are several methods for finding the equation of a tangent line given x, such as using the point-slope form or the slope-intercept form. The method used may vary depending on the given information and the complexity of the function.

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