# Math(Please check)

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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) -

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

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