Posted by Hannah on .
Find the slope of the graph at
y=sqrt 2x^2 +1 at the point (2,3).
First I plugged 2 into the equation and got 3. Then the derivative of the equation is 4x and then I multiplied that by 3 and got 12. Is this correct?

Math(Please check) 
helper,
y=sqrt 2x^2 +1 at the point (2,3).
No, you are not correct.
First, you need to find the derivative of,
(sqrt(2x^2 + 1))
dy/dx = 2x(2x^2 + 1)^(1/2)
This is the slope of the tangent line at P.
Now plug in P(2,3)
2x(2x^2 + 1)^(1/2)
2(2)(2(2^2) + 1)^(1/2)
4(8 + 1)^(1/2)
4 * (9)^(1/2) = 4 * 1/(9^(1/2))= 4/3