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I can't figure out how to evaluate a partial derivative of the form

$$\frac{\partial F(x,y(x),z(x))}{\partial x}$$

I know that if it was

$$\frac{\partial F(x,y,z)}{\partial x}$$

Then we differentiate as normal but taking $y$ and $z$ as constant. But $y$ and $z$ depend on $x$ so I guess this won't work.

I also know that if it was $$\frac{\partial F(y(x),z(x))}{\partial x}$$ Then we use the chain rule to get $$\frac{\partial F(y(x),z(x))}{\partial x}=\frac{\partial F}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}$$

But Im unsure to how to solve my original problem.

Connor Bishop
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    hint: use your last formula for the chain rule with 3 $(x,y,z)$ instead of 2 $(y,z)$ variables and then use $\frac{\partial x}{\partial x}=1$ – Max May 01 '16 at 09:04
  • But wouldn't I then get $\frac{\partial F}{\partial x}=\frac{\partial F}{\partial x}$+more terms – Connor Bishop May 01 '16 at 09:34

1 Answers1

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You are composing a function $$\phi:\quad{\mathbb R}\to{\mathbb R}^3,\qquad x\mapsto\bigl(x,y(x),z(x)\bigr)$$ with an $$F:\quad{\mathbb R}^3\to{\mathbb R},\qquad (x_1,x_2,x_3)\mapsto F(x_1,x_2,x_3)\ ,$$ so that you are looking at $$f(x):=F\bigl(\phi(x)\bigr)=F(x,y(x),z(x)\bigr)\ .$$ By the chain rule you get $$f'(x)={\partial F\over\partial x_1}\biggr|_{\phi(x)}\cdot 1+{\partial F\over\partial x_2}\biggr|_{\phi(x)}\cdot y'(x)+{\partial F\over\partial x_3}\biggr|_{\phi(x)}\cdot z'(x)\ ,$$ which in your context is written as $$f'(x)={\partial F\over\partial x}+{\partial F\over\partial y}\cdot y'(x)+{\partial F\over\partial z}\cdot z'(x)\ .$$

Christian Blatter
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