Hello All,

Would someone mind helping me check my work? My answer is very close to the answer given in the back of the book, but isn't exactly the same.

$\displaystyle y=3x^2+\sqrt{2}$

$\displaystyle z=\sqrt{1+y}$

$\displaystyle v=\frac{1}{\sqrt{3}+4z}$ find $\displaystyle \frac{dv}{dx}$

$\displaystyle \frac{dy}{dx}=$ $\displaystyle 3x$

$\displaystyle \frac{dz}{dy}=$ $\displaystyle \frac{1}{2\sqrt{1+y}}$

$\displaystyle \frac{dv}{dz}=$ $\displaystyle \frac{-4}{(\sqrt{3}+4z)^2}$

$\displaystyle \frac{dy}{dx}*\frac{dz}{dy}*\frac{dv}{dz}=\frac{dv }{dx}$

$\displaystyle 3x$ * $\displaystyle \frac{1}{2\sqrt{1+y}}$ * $\displaystyle \frac{-4}{(\sqrt{3}+4z)^2}$

3x * $\displaystyle \frac{1}{2\sqrt{1+(3x^2+\sqrt{2})}}$ * $\displaystyle \frac{-4}{(\sqrt{3}+4\sqrt{1+3x^2+\sqrt{2})^2}}$

Here is the messy answer I ended up with. The answer in the back of the book matches this exactly EXCEPT it is missing the "2" in the beginning of the denominator...any thoughts?

$\displaystyle \frac{-12x}{2\sqrt{1+3x^2+\sqrt{2}}(\sqrt{3}+4\sqrt{1+3x^ 2+\sqrt{2}})^2}$