If y2=4ax, then dydx is - Infinity Learn

A
$\sqrt{\frac{a}{x}}$
B
$\sqrt{a x}$
C
$\sqrt{a + x}$
D
$\sqrt{a - x}$

Solution:

$\begin{array}{l} \frac{d}{d x} (y^{2}) = \frac{d}{d x} (4 a x) \Rightarrow 2 y . \frac{d y}{d x} = 4 a \frac{d}{d x} (x) \\ \Rightarrow 2 y \frac{d y}{d x} = 4 a (1) \Rightarrow \frac{d y}{d x} = \frac{4 a}{2 y} \\ \frac{d y}{d x} = \frac{2 a}{y} = \frac{2 a}{\sqrt{4 a x}} = \sqrt{\frac{a}{x}} \end{array}$

If $y^{2} = 4 a x$ , then $\frac{d y}{d x}$ is

Solution:

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