I need Calculus help!!!!?

Question

Find the slope of the tangent line to the graph of the function f(x) = sq. root symbol 3x^2 + 6 at the point (1, 3)

&

Find the derivative of the function f(x) = sq. root symbol 3x^2 - x

I have a test coming up and I'm having trouble with these two problems. Help would be appreciated!

Answer 1

1.) f(x) = √(3x² + 6) = (3x² + 6)^(1/2)

To find the slope of the tangent line to the graph at a certain point you need to find the derivative of he function, then plug in the x - coordinate of the point to get the exact value:

f'(x) = (1/2) · (3x² + 6)^(-1/2) · (6x) = 3x / √(3x² + 6)
f'(1) = 3·1 / √(3·1² + 6) = 3 / √9 = 1

The slope is 1.


2.) f(x) = √(3x² - x) = (3x² - x)^(1/2)

f'(x) = (1/2) · (3x² - x)^(-1/2) · (6x - 1) = (6x-1) / 2√(3x² - x)

Hope this helps!

Answer 2

To find the slope of the tangent line at any point, you must first solve for the first derivative.

F(x) sqrt(3x^2 +6) = (3x^2 +6)^ (1/2)
using the power rule and chain rule
F'(x) = (1/2) [(3x^2 +6)^(-1/2)] * (6x)
= 6x/ (sqrt(3x^2 +6)

to find the slope at (1,3) subsitute the x coordinate of 1 into the F'(x) function which results in:
6/9 = 2/3



Part two is similar.
f(x)= (3x^2 - x)^(1/2)
f'(x) = (1/2)[(3x^2 - x)^(-1/2)] * (6x-1)
simplifies to
(6x-1) / [2(3x^2-x)^(1/2)]

Answer 3

y = (3x^2 + 6)^(1/2)
y' = (1/2)(6x)(3x^2 + 6)^-(1/2)
y'(1) = 3/3 = 1

Switch to exponential notation and use the power and composite rules:
f(x) = (3x^2 - x)^(1/2)
let u = 3x^2 - x
Then f(u) = u^(1/2) and
f'(u) = (1/2)u^-(1/2) du/dx
(d/dx[u^n] = nu^(n - 1) )
but du/dx = 6x - 1, so
f'(x) = (1/2)(6x - 1)(3x^2 - x)^-(1/2)