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I have the function $u=\ln(1+x+y^2-z^2)$. I have to find the value of the expression $u'(x) +u'(y) +u'(z)$ for $z=y=x=1$. In a usual case, I keep one of the variables constant, how do I keep two here?

haunted85
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Tweetingbird
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    In doing a partial, all other vars are constant. The notation for partial w.r.t. $x$ is usually $\partial/\partial x$, not $u'(x)$ -- the "prime" notation is usually for derivative in the one variable case. – coffeemath Mar 26 '13 at 15:09
  • Let $a\in \Bbb R$. You know how to find the derivative of $x\mapsto \ln(1+x+a^2)$, right? Now take $a\in \Bbb R,b\in \Bbb R$. You still know how to find the derivative of $x\mapsto \ln(1+x+a^2-b^2)$, right? Well you're asked to exactly that. All the variables except the one with respect to which you want to find the partial derivative are arbitrary constants that you can think of as chosen before defining your function. – xavierm02 Mar 26 '13 at 15:11
  • Instead of considering $(x,y,z)\mapsto \ln(1+x+y^2-z^2)$, you consider $y,z\in \Bbb R$ and $x\mapsto \ln(1+x+y^2-z^2)$ – xavierm02 Mar 26 '13 at 15:13

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For example with the partial derivative with respect to $x$ just write $$u = \ln((1+y^2-z^2)+x).$$ Then treat $(1+y^2-z^2)$ as a constant.

Michael Albanese
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  • @Tweetingbird I think Joe is right there must be a mistake in your book. –  Mar 26 '13 at 16:31
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Treat both variable as one large constant. Something like $\frac{\partial}{\partial x}(1+y^2-z^2)=0$. $$u_x=\frac{1}{1+x+y^2-z^2}(1)$$ (using chain rule). $$u_y=\frac{1}{1+x+y^2-z^2}(2y)$$ $$u_z=\frac{1}{1+x+y^2-z^2}(-2z)$$ $$u_{x}(1)+u_{y}(1)+u_{z}(1)=\frac{1}{2}(1+2-2)=\boxed{\frac{1}{2}}$$.

Joe
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