Limits and Derivatives help?

Question

This is the only problem giving me trouble in this section.


Find an equation of the line tangent to the curve y=x+(1/x) at the point (5,26/5).

Answer 1

f `(x) = 1 - 1 / x ²
f `(5) = 1 - 1 / 25
f `(5) = 24 / 25 = m
y - 26/5 = (24/25) (x - 5)
y = (24/25)x - 24/5 + 26/5
y = (24/25)x + 2/5 is required tangent line.
If desired, may be written as:-
25y = 24x + 10
24x - 25y + 10 = 0

Answer 2

First, compute the derivative. This will give you the slope of the tangent line.

y = x + (1/x)
y = x + x^(-1)
y' = 1 + (-1)x^(-2)
y' = 1 - 1/x^2

Now plug the x-value of your point (5) into the equation for y':

y'(5) = 1 - 1/5^2 = 1 - 1/25 = 24/25.

So the slope of the tangent line is 24/25.

The tangent line also passes through (5, 26/5), so by the point-slope form of a line, we have

y - 26/5 = 24/25(x - 5).

Answer 3

The derivative of x + (1/x) is 1 - 1/x^2.
Stick x = 5 in that and you get 1 - 1/25 or 24/25

That's the slope of your tangent line. From here on, it's analytic geometry.

(y - y1) = m(x - x1)
(y - 26/5) = 24/25(x - 5)
y - 26/5 = 24/25 * x - 24/5
y - 24/25 * x = 2/5
-24x + 25y = 10

Answer 4

First, find the derivative of the equation.

y = x + x^-1, so y' = 1 - x^-2

Now find the slope at the point x = 5: 1 - 5^-2 = 24/25.

Finally, use point-slope. y - y1 = m(x - x1) -->

y - 26/5 = (24/25)(x - 5) --> y = (24/25)x + 2/5.

Answer 5

y= x+ 1/x

dy/dx=1- 1/x2 (x-square)
then [dy/dx ] at the point [5,26/5]=1-1/25=24/25

so the equation is written as,

y - y1 (y 0ne) = [dy/dx] at the point [x1,y1] * (x - x1)

that is,

y -26/5 = 24/25 * (x -5)

or, 24x -25y +10=0 (ans:)

reply me this is right or not.

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Answer 6

y' = 1 - 1/x^2
y'(5) = 1 - 1/25
y'(5) = 24/25 ---> slope

(y - 26/5) = 24/25 (x - 5)
y = (24/25)x - 24/5 + 26/5
y = (24/25)x + 2/5

Answer 7

dy/dx = 1 - 1/x^2