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y=sqrt(3-sqrt(x)) y'=?

 Jun 25, 2019

Best Answer 

 #1
avatar+8842 
+4

\(y\ =\ \sqrt{3-\sqrt{x}}\\~\\ y\ =\ (3-x^\frac12)^\frac12\\~\\ y'\ =\ \frac{d}{dx}(3-x^\frac12)^\frac12\)

 

Using the power rule and chain rule,

 

\(y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ \frac{d}{dx}(3-x^\frac12)\ )\\~\\ y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ \frac{d}{dx}3-\frac{d}{dx}x^\frac12\ )\\~\\ y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ 0-\frac12x^{-\frac12}\ )\)

 

And we can rewrite the right side of the equation like this:

 

\( y'\ =\ \dfrac{-1}{4\,\cdot\,\sqrt{3-\sqrt{x}}\,\cdot\,\sqrt{x}} \)_

 Jun 25, 2019
 #1
avatar+8842 
+4
Best Answer

\(y\ =\ \sqrt{3-\sqrt{x}}\\~\\ y\ =\ (3-x^\frac12)^\frac12\\~\\ y'\ =\ \frac{d}{dx}(3-x^\frac12)^\frac12\)

 

Using the power rule and chain rule,

 

\(y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ \frac{d}{dx}(3-x^\frac12)\ )\\~\\ y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ \frac{d}{dx}3-\frac{d}{dx}x^\frac12\ )\\~\\ y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ 0-\frac12x^{-\frac12}\ )\)

 

And we can rewrite the right side of the equation like this:

 

\( y'\ =\ \dfrac{-1}{4\,\cdot\,\sqrt{3-\sqrt{x}}\,\cdot\,\sqrt{x}} \)_

hectictar Jun 25, 2019
 #2
avatar+23862 
+3

y=\(\sqrt{3-\sqrt{x}} \)

\(y'=\ ?\)

 

\(\begin{array}{|rcll|} \hline y &=& \sqrt{3-\sqrt{x}} \\ y^2 &=& 3-\sqrt{x} \quad | \quad \text{derivate both sides} \\\\ 2yy' &=& -\dfrac{d\ ( \sqrt{x})}{dx} \\\\ 2yy' &=& -\dfrac{d\ (x^{\frac12})}{dx} \\\\ 2yy' &=& -\dfrac{1}{2} x^{-\frac{1}{2}} \\\\ 2yy' &=& -\dfrac{1}{2x^{\frac{1}{2}}} \\\\ 2yy' &=& -\dfrac{1}{2\sqrt{x}} \\\\ y' &=& -\dfrac{1}{2\sqrt{x}2y} \\\\ y' &=& -\dfrac{1}{4\sqrt{x} y} \\\\ \mathbf{y'} &=& \mathbf{-\dfrac{1}{4\sqrt{x} \sqrt{3-\sqrt{x}}}} \\ \hline \end{array}\)

 

laugh

 Jun 26, 2019

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