Question

a Find, from first principles, the derivatives of (3x+5)/x^(1/2)

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Answer 1

Step-by-step explanation:

We have,

$y = \frac{3x + 5}{ \sqrt{x} }$

$= > \frac{dy}{dx} = \lim_{h\rarr0 } \frac{f(x + h) - f(x)}{h}$

$= > \frac{dy}{dx} = \lim_{h\rarr0 } \frac{ \frac{3(x + h) + 5}{ \sqrt{x + h} } - \frac{3x + 5}{\sqrt{x} } }{h}$

$= > \frac{dy}{dx} = \lim_{h\rarr0 } \frac{(3x + 3h + 5) \sqrt{x} - (3x + 5) \sqrt{x + h} }{h \sqrt{x(x + h)} }$

$= > \frac{dy}{dx} = \lim_{h\rarr0 } \frac{(3x + 5)( \sqrt{x} - \sqrt{x + h} ) + 3h \sqrt{x} }{h \sqrt{x(x + h)} }$

$= > \frac{dy}{dx} = \lim_{h\rarr0 } \frac{(3x + 5)( - h)}{h \sqrt{x(x + h)} ( \sqrt{x} + \sqrt{x + h} )} + \frac{3 \sqrt{x} }{x}$

$= > \frac{dy}{dx} = - \lim_{h\rarr0 } \frac{(3x +5)}{ \sqrt{x(x + h)} ( \sqrt{x} + \sqrt{x + h} )} + \frac{3}{ \sqrt{x} }$

$= > \frac{dy}{dx} = - \frac{(3x +5)}{2 x\sqrt{x} } + \frac{3}{ \sqrt{x} }$