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

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

Answer 1

Emilia Aguilar

Your answer is right

We are given #g(y)=sqrt[2y+(3y+4y^2)^3]=(2y+(3y+4y^2)^3)^(1/2)#

#g(y)=d/dy[(2y+(3y+4y^2)^3)^(1/2)]# #=1/2*(2y+(3y+4y^2)^3)^(1/2-1)* d/dy[2y+(3y+4y^2)^3]#

#=(2y+(3y+4y^2)^3)^(-1/2)/2* (d/dy[2y]+d/dy[(3y+4y^2)^3])#

#=(2y+(3y+4y^2)^3)^(-1/2)/2* (2+(3*(3y+4y^2)^(3-1)d/dy[3y+4y^2]))#

#=(2y+(3y+4y^2)^3)^(-1/2)/2* (2+3(3y+4y^2)^2(3+8y))#

#=((2y+(3y+4y^2)^3)^(-1/2)(2+3(3y+4y^2)^2(3+8y)))/2#

#=(2+3(3y+4y^2)^2(3+8y))/(2(2y+(3y+4y^2)^3)^(1/2))#

#=(2+(9+24y)(3y+4y^2)^2)/(2sqrt(2y+(3y+4y^2)^3))#

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Answer 2

Charles Arocha

To find the derivative of ( g(y) = \sqrt{2y + (3y + 4y^2)^3} ), use the chain rule:

[ g'(y) = \frac{1}{2\sqrt{2y + (3y + 4y^2)^3}} \times \left(2 + 9(3y + 4y^2)^2\right) ]

Simplify this expression as needed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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

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