Thread: Complex Chain Rule Problem

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?

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

mistake is, must be
dy/dx=6x

Thank You!