Equation of a tangent line given x

[Drewnlauren06]

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

 

[tiny-tim]

Welcome to PF!

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

Hi Drewnlauren06! Welcome to PF!

Do you know how to find the slope (the y/x) of the tangent line?​

 

[Drewnlauren06]

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.

 

[tiny-tim]

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? ​

 

[Drewnlauren06]

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.

 

[tiny-tim]

Hi Drewnlauren06!

(have a square-root: √ and use the X² tag, just above the reply box )

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

… 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