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?
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
To find the slope of the graph at a specific point, you need to take the derivative of the given equation and then evaluate it at that point.
Let's start by finding the derivative of the equation y = √(2x^2 + 1). To do this, you can use the power rule and the chain rule. The power rule states that the derivative of x^n with respect to x is nx^(n-1), and the chain rule states that the derivative of f(g(x)) with respect to x is f'(g(x)) * g'(x).
Using the chain rule, we can rewrite the equation as y = (2x^2 + 1)^(1/2). Applying the power rule, the derivative of this equation with respect to x is:
dy/dx = (1/2)(2x^2 + 1)^(-1/2) * d/dx(2x^2 + 1).
Now, let's find d/dx(2x^2 + 1):
d/dx(2x^2 + 1) = 4x.
Substituting this back into the derivative equation, we get:
dy/dx = (1/2)(2x^2 + 1)^(-1/2) * 4x.
Now, to find the slope at the point (2,3), we substitute x = 2 into the derivative equation:
dy/dx = (1/2)(2(2)^2 + 1)^(-1/2) * 4(2)
= (1/2)(2(4) + 1)^(-1/2) * 8
= (1/2)(8 + 1)^(-1/2) * 8
= (1/2)(9)^(-1/2) * 8
= (1/2)(3)^(-1) * 8
= (1/2)(1/3) * 8
= (1/6) * 8
= 8/6
= 4/3.
Therefore, the slope of the graph at the point (2,3) is 4/3.