Question

1.If f(X)=4-2x/2+3x+3xpower2 then the values of x for which f'(X)=0 is
2.If y=(x+square root of x power 2 -1)power m then(x power 2 -1)(dy/dx)power 2-mpower2 y power2 is proved to be

Answer 1

Answer on Question#54571, Math, Differential Equations

1. $f'(x) = \frac{(-2)(2 + 3x + 3x^2) - (3 + 6x)(4 - 2x)}{(2 + 3x + 3x^2)^2} = \frac{6x^2 - 24x - 16}{(2 + 3x + 3x^2)^2} = 0$. Solving the quadratic equation in the numerator, obtain $x_{1,2} = \frac{2}{3}\left(3 \pm \sqrt{15}\right)$. These are the values for which $f'(x) = 0$.

2. $y = \left(x + \left(\sqrt{x}\right)^{2} - 1\right)^{m} = (2x - 1)^{m}$.

$\frac{dy}{dx} = m(2x - 1)^{-1 + m}(2) = 2m[2x - 1]^{m - 1}$, therefore

$$\begin{array}{l} (x^{2} - 1)\left(\frac{dy}{dx}\right)^{2} - m^{2}y^{2} = 4m^{2}(x^{2} - 1)(2x - 1)^{2m - 2} - m^{2}(2x - 1)^{2m} = \\ = m^{2}(2x - 1)^{2m - 2}[4x^{2} - 4 - (2x - 1)^{2}] = m^{2}(2x - 1)^{2m - 2}[-5 + 4x]. \end{array}$$

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