MHB How to Find the Second Derivative with Given Equation at a Specific Point?

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To find the second derivative at the point (3,0) for the equation (x+2y)·(dy/dx)=2x-y, first calculate dy/dx using the formula dy/dx=(2x-y)/(x+2y). After substituting the point into this equation, the first derivative is determined. Then, apply implicit differentiation to find d^2y/dx^2 by differentiating dy/dx again. The final result provides the value of the second derivative at the specified point.
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If $(x+2y)\cdot \dfrac{dy}{dx}=2x-y$ what is the value of $\dfrac{d^2y}{dx^2}$ at the point (3,0)?
ok not sure of the next step but
$\dfrac{dy}{dx}=\dfrac{2x-y}{x+2y}$
 
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Re: 231 value of second dirivative

See https://mathhelpboards.com/calculus-10/297-ap-calculus-exam-2nd-derivative-26690.html.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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