But if were dealing with functions that take a point in mathbb Rn as input, then were just doing vector calculus. If were dealing with functions that take a matrix as input, then were doing matrix calculus. Show Solution For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. begingroup Hmm, I think the key phrase in the first sentence of that article is 'especially over spaces of matrices'. Hardy, ``A course of Pure Mathematics,'' Cambridge University Press, 1960, 10th Edition, p. Hint : Recall that with Chain Rule problems you need to identify the inside and outside functions and then apply the chain rule. Even though there is some of the same 'canceling' trick, the equation doesn't quite make as much. Even though there is some of the same 'canceling' trick, the equation doesnt quite make as much. Consider z f(x(t), y(t)) z f ( x ( t), y ( t)), then its chain rule derivative is: dz dt f x dx dt + f y dy dt d z d t f x d x d t + f y d y d t. Consider z f(x(t), y(t)) z f ( x ( t), y ( t)), then its chain rule derivative is: dz dt f x dx dt + f y dy dt d z d t f x d x d t + f y d y d t. However, for multiple variables the equation looks very different. One approach is to use the fact the "differentiability" is equivalent to "approximate linearity", in the sense that if $f$ is defined in some neighborhood of $a$, thenį'(a) = \lim_&\rightarrow 0 = F'(y)\,f'(x) However, for multiple variables the equation looks very different.
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