How do you find the derivative of $y = {(3 x + 4 x - x^{2})}^{\frac{1}{2}}$ $y=(3x+4x-x^2)^(1/2)$ ?

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How do you find the derivative of $y = {(3 x + 4 x - x^{2})}^{\frac{1}{2}}$ $y=(3x+4x-x^2)^(1/2)$ ?

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Answer 1 · 2016-11-13T06:29:53

$\frac{2 - x}{\sqrt{3 + 4 x - x^{2}}}$ $(2-x)/(sqrt(3+4x-x^2))$

Explanation:

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${(3 + 4 x - x^{2})}^{\frac{1}{2}}$ $(3+4x-x^2)^(1/2)$
I will be going off of that one for now.

Calculation

Apply chain rule:
$\frac{1}{2} {(3 + 4 x - x^{2})}^{- \frac{1}{2}} \times (4 (1) - 2 x)$ $1/2(3+4x-x^2)^{-1/2}\times(4(1)-2x)$
$= \frac{1}{2 {(3 + 4 x - x^{2})}^{\frac{1}{2}}} \times (4 - 2 x)$ $=1/(2(3+4x-x^2)^(1/2)]\times(4-2x)$
$= \frac{4 - 2 x}{2 {(3 + 4 x - x^{2})}^{\frac{1}{2}}}$ $=(4-2x)/(2(3+4x-x^2)^(1/2)$
$= \frac{2 (2 - x)}{2 {(3 + 4 x - x^{2})}^{\frac{1}{2}}}$ $=(2(2-x))/(2(3+4x-x^2)^(1/2))$
$= \frac{2 - x}{\sqrt{3 + 4 x - x^{2}}}$ $=(2-x)/(sqrt(3+4x-x^2))$

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How do you find the derivative of $y = {(3 x + 4 x - x^{2})}^{\frac{1}{2}}$ $y=(3x+4x-x^2)^(1/2)$ ?

1 Answer

Explanation:

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Calculation

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How do you find the derivative of $y = {(3 x + 4 x - x^{2})}^{\frac{1}{2}}$ $y=(3x+4x-x^2)^(1/2)$ ?