Question

find an equation of the tangent line to the graph of f at the given point.
f(x)=x√x^2+5) point (2,6)

Answer 1

i have to use the product rule, right?

Answer 2

Nah you got to use me

Answer 3

a trinity rule? o.0

Answer 4

y = x √(x^2 + 5)
=> dy/dx = √(x^2 + 5) + x * x / √(x^2 + 5)
=> dy/dx at (2, 6)
= √(2^2 + 5) + 2^2 / √(2^2 + 5)
= 3 + 4/3
= 13/3
The equation of the tangent line passes through (2, 6) and has a slope = 13/3
=> the eqn. of the tangent line is
y - 6 = (13/3) (x - 2)
=> 13x - 3y = 8.

Answer 5

Trinity does rule =)

Answer 6

that was fast.

Answer 7

Like I said,I am trinity

Answer 8

Need other help?

Answer 9

in a few minutes, yea. just reading over your answer

Answer 10

No doubt

Answer 11

so how did you know to get the equation, => dy/dx = √(x^2 + 5) + x * x / √(x^2 + 5)

Answer 12

You find the square root f:(X²+ 5))

Answer 13

product rule i think

Answer 14

Post your other question?

Answer 15

I might fall asleep waiting lol

Answer 16

oh sorry, lol. these questions you can;t help me with :( i have to look at these graphs. thanks though. goodnight!

Answer 17

;o