How do you differentiate $x^{\sqrt{5}} + \sqrt{5 x}$ $x^sqrt5+sqrt(5x)$ ?

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Answer 1 · 2017-11-25T17:05:00

$\frac{d y}{d x} = \sqrt{5} \cdot x^{\sqrt{5} - 1} + \frac{\sqrt{5}}{2 \sqrt{x}} .$ $dy/dx=sqrt5*x^(sqrt5-1)+sqrt5/(2sqrtx).$

$\frac{d}{d x} {x^{\sqrt{5}} + \sqrt{5 x}},$ $d/dx{x^(sqrt5)+sqrt(5x)},$

$= \frac{d}{d x} {x^{\sqrt{5}}} + \frac{d}{d x} {\sqrt{5} \cdot x^{\frac{1}{2}}},$ $=d/dx{x^sqrt5}+d/dx{sqrt5*x^(1/2)},$

$= \sqrt{5} \cdot x^{\sqrt{5} - 1} + \sqrt{5} \cdot \frac{d}{d x} {x^{\frac{1}{2}}},$ $=sqrt5*x^(sqrt5-1)+sqrt5*d/dx{x^(1/2)},$

$= \sqrt{5} \cdot x^{\sqrt{5} - 1} + \sqrt{5} {\frac{1}{2} x^{(\frac{1}{2}) - 1}},$ $=sqrt5*x^(sqrt5-1)+sqrt5{1/2x^((1/2)-1)},$

$= \sqrt{5} \cdot x^{\sqrt{5} - 1} + \sqrt{5} {\frac{1}{2} x^{- \frac{1}{2}}} .$ $=sqrt5*x^(sqrt5-1)+sqrt5{1/2x^(-1/2)}.$

$\Rightarrow \frac{d y}{d x} = \sqrt{5} \cdot x^{\sqrt{5} - 1} + \frac{\sqrt{5}}{2 \sqrt{x}} .$ $rArr dy/dx=sqrt5*x^(sqrt5-1)+sqrt5/(2sqrtx).$

How do you differentiate x^sqrt5+sqrt(5x)x√5+√5xx^sqrt5+sqrt(5x)?