What is the derivative of $\sqrt{2 x}$ $sqrt(2x)$ ?

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Answer 1 · 2014-12-13T13:54:28

Power rule: $\frac{d y}{d x} [x^{n}] = n \cdot x^{n - 1}$ $(dy)/(dx)[x^n]=n*x^(n-1)$

Power rule + chain rule: $\frac{d y}{d x} [u^{n}] = n \cdot u^{n - 1} \cdot \frac{d u}{d x}$ $(dy)/(dx)[u^n]=n*u^(n-1)*(du)/(dx)$

Let $u = 2 x$ $u=2x$ so $\frac{d u}{d x} = 2$ $(du)/(dx)=2$

We're left with $y = \sqrt{u}$ $y=sqrt(u)$ which can be rewritten as $y = u^{\frac{1}{2}}$ $y=u^(1/2)$

Now, $\frac{d y}{d x}$ $(dy)/(dx)$ can be found using the power rule and the chain rule.

Back to our problem: $\frac{d y}{d x} = \frac{1}{2} \cdot u^{- \frac{1}{2}} \cdot \frac{d u}{d x}$ $(dy)/(dx)= 1/2 * u^(-1/2)*(du)/(dx)$

plugging in $\frac{d u}{d x}$ $(du)/(dx)$ we get:

$\frac{d y}{d x} = \frac{1}{2} \cdot u^{- \frac{1}{2}} \cdot (2)$ $(dy)/(dx)= 1/2 * u^(-1/2)*(2)$

we know that: $\frac{2}{2} = 1$ $2/2=1$

therefore, $\frac{d y}{d x} = u^{- \frac{1}{2}}$ $(dy)/(dx)=u^(-1/2)$

Plugging in the value for $u$ $u$ we find that:
$\frac{d y}{d x} = 2 x^{- \frac{1}{2}}$ $(dy)/(dx)=2x^(-1/2)$

What is the derivative of sqrt(2x)√2xsqrt(2x)?