What is the derivative: $g (y) = \sqrt{2 y + {(3 y + 4 y^{2})}^{3}}$ $g(y) = sqrt[2y + (3y + 4y^2)^3]$ ?

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What is the derivative: $g (y) = \sqrt{2 y + {(3 y + 4 y^{2})}^{3}}$ $g(y) = sqrt[2y + (3y + 4y^2)^3]$ ?

I got to
$\frac{2 + (9 + 24 y) {(3 y + 4 y^{2})}^{2}}{2 \sqrt{2 y + {(3 y + 4 y^{2})}^{3}}}$ $[2 + (9 + 24y)(3y + 4y^2)^2]/{2 sqrt [2y + (3y + 4y^2)^3]}$ .

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Answer 1 · 2018-02-07T13:48:38

Your answer is right

Explanation:

We are given $g (y) = \sqrt{2 y + {(3 y + 4 y^{2})}^{3}} = {(2 y + {(3 y + 4 y^{2})}^{3})}^{\frac{1}{2}}$ $g(y)=sqrt[2y+(3y+4y^2)^3]=(2y+(3y+4y^2)^3)^(1/2)$

$g (y) = \frac{d}{d y} [{(2 y + {(3 y + 4 y^{2})}^{3})}^{\frac{1}{2}}]$ $g(y)=d/dy[(2y+(3y+4y^2)^3)^(1/2)]$
$= \frac{1}{2} \cdot {(2 y + {(3 y + 4 y^{2})}^{3})}^{\frac{1}{2} - 1} \cdot \frac{d}{d y} [2 y + {(3 y + 4 y^{2})}^{3}]$ $=1/2*(2y+(3y+4y^2)^3)^(1/2-1)* d/dy[2y+(3y+4y^2)^3]$

$= \frac{{(2 y + {(3 y + 4 y^{2})}^{3})}^{- \frac{1}{2}}}{2} \cdot (\frac{d}{d y} [2 y] + \frac{d}{d y} [{(3 y + 4 y^{2})}^{3}])$ $=(2y+(3y+4y^2)^3)^(-1/2)/2* (d/dy[2y]+d/dy[(3y+4y^2)^3])$

$= \frac{{(2 y + {(3 y + 4 y^{2})}^{3})}^{- \frac{1}{2}}}{2} \cdot (2 + (3 \cdot {(3 y + 4 y^{2})}^{3 - 1} \frac{d}{d y} [3 y + 4 y^{2}]))$ $=(2y+(3y+4y^2)^3)^(-1/2)/2* (2+(3*(3y+4y^2)^(3-1)d/dy[3y+4y^2]))$

$= \frac{{(2 y + {(3 y + 4 y^{2})}^{3})}^{- \frac{1}{2}}}{2} \cdot (2 + 3 {(3 y + 4 y^{2})}^{2} (3 + 8 y))$ $=(2y+(3y+4y^2)^3)^(-1/2)/2* (2+3(3y+4y^2)^2(3+8y))$

$= \frac{{(2 y + {(3 y + 4 y^{2})}^{3})}^{- \frac{1}{2}} (2 + 3 {(3 y + 4 y^{2})}^{2} (3 + 8 y))}{2}$ $=((2y+(3y+4y^2)^3)^(-1/2)(2+3(3y+4y^2)^2(3+8y)))/2$

$= \frac{2 + 3 {(3 y + 4 y^{2})}^{2} (3 + 8 y)}{2 {(2 y + {(3 y + 4 y^{2})}^{3})}^{\frac{1}{2}}}$ $=(2+3(3y+4y^2)^2(3+8y))/(2(2y+(3y+4y^2)^3)^(1/2))$

$= \frac{2 + (9 + 24 y) {(3 y + 4 y^{2})}^{2}}{2 \sqrt{2 y + {(3 y + 4 y^{2})}^{3}}}$ $=(2+(9+24y)(3y+4y^2)^2)/(2sqrt(2y+(3y+4y^2)^3))$

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What is the derivative: $g (y) = \sqrt{2 y + {(3 y + 4 y^{2})}^{3}}$ $g(y) = sqrt[2y + (3y + 4y^2)^3]$ ?

I got to
$\frac{2 + (9 + 24 y) {(3 y + 4 y^{2})}^{2}}{2 \sqrt{2 y + {(3 y + 4 y^{2})}^{3}}}$ $[2 + (9 + 24y)(3y + 4y^2)^2]/{2 sqrt [2y + (3y + 4y^2)^3]}$ .

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Explanation:

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What is the derivative: g(y) = sqrt[2y + (3y + 4y^2)^3]g(y)=√2y+(3y+4y2)3g(y) = sqrt[2y + (3y + 4y^2)^3]?

I got to [2 + (9 + 24y)(3y + 4y^2)^2]/{2 sqrt [2y + (3y + 4y^2)^3]}2+(9+24y)(3y+4y2)22√2y+(3y+4y2)3[2 + (9 + 24y)(3y + 4y^2)^2]/{2 sqrt [2y + (3y + 4y^2)^3]}.

1 Answer

Explanation:

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Impact of this question

What is the derivative: $g (y) = \sqrt{2 y + {(3 y + 4 y^{2})}^{3}}$ $g(y) = sqrt[2y + (3y + 4y^2)^3]$ ?

I got to
$\frac{2 + (9 + 24 y) {(3 y + 4 y^{2})}^{2}}{2 \sqrt{2 y + {(3 y + 4 y^{2})}^{3}}}$ $[2 + (9 + 24y)(3y + 4y^2)^2]/{2 sqrt [2y + (3y + 4y^2)^3]}$ .