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If x = t^2 + 1 and y = t^3, then d^2y/dx^2 = I know I can solve for t in terms of x and substitute that into y = t^3 and find the double derivative. I also know that I can take the derivative of x and y then divide dy/dt by dx/dt.
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R=M^2(c/2m/3) dR/dM=CMM^2 I found the derivative. Now how would I find the vale of M that maximize the derivative dR/dM? set it to zero, and solve for m. You get two solutions. Use the second derivative to see which one is the
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Also I would like to know how to set this problem up in finding the answer: Find the second derivative. y= sqrt(3x+4) Thanks, P.S. I tried taking 3x+4 and squaring it to the 1/2. I thought that might've been the same, then I took
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Suppose f(x) and g(x) are functions of x differentiable at x = 4. Given that f(4) = 3, f'(4) = 2, g(4) = 6, g'(4) = 4, find the value of each of the following. a. Derivative of f(x) * g(x) My answer: 27 b. Derivative of
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Suppose f(x) and g(x) are functions of x differentiable at x = 4. Given that f(4) = 3, f'(4) = 2, g(4) = 6, g'(4) = 5, find the value of each of the following. a. Derivative of f(x) * g(x) My answer: 27 b. Derivative of
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Find the first and second derivative  simplify your answer. y=x/4x+1 I solved the first derivative and got 1/(4x+1)^2 Not sure if I did the first derivative right and not sure how to do the second derivative.
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