Complex Chain Rule Problem

**dbakeg00** · Mar 31st 2010, 01:36 PM

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.

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

$z=\sqrt{1+y}$

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

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

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

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

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

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

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

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?

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

**zzzoak** · Mar 31st 2010, 01:49 PM

mistake is, must be
dy/dx=6x

**dbakeg00** · Mar 31st 2010, 01:53 PM

Thank You!

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