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The following question is regarding vector calculus. Say that I have a vector $\pmb A$ and a scalar $\lambda$. Does the following relationship hold true for $\frac{d\pmb A}{d\lambda}$ and $\frac{d\lambda}{d\pmb A}$. Let, $$\pmb A = (a_1 \space a_2 \space a_3 \space ...a_n)^T$$ Then, $$\frac{d\pmb A}{d\lambda} = (\frac{da_1}{d\lambda} \space \frac{da_2}{d\lambda} \space \frac{da_3}{d\lambda} \space ...\frac{da_n}{d\lambda})^T$$ and, $$\frac{d\lambda}{d\pmb A} = (\frac{d\lambda}{da_1} \space \frac{d\lambda}{da_2} \space \frac{d\lambda}{da_3} \space ...\frac{d\lambda}{da_n})$$ Is the relationship, $$\frac{d\lambda}{d\pmb A} = (\frac{1}{{da_1} \over {d\lambda}} \space \frac{1}{{da_2} \over {d\lambda}} \space \frac{1}{{da_3} \over {d\lambda}} \space ...\frac{1}{{da_n} \over {d\lambda}} )$$ valid still ?

(Here, $\frac{da_i}{d\lambda} \neq 0 $ for $i = 1, 2, ..., n$).

makmd
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    What is the meaning of $dA/d\lambda$? What about the more weird of $d\lambda/dA$? If you could answer me, probably you know the answer. – azif00 Apr 07 '20 at 06:00
  • It is valid because each component of $\mathbf{a}$ is a scalar function of a single variable and that relationship with derivatives for single variable, invertible (nonzero derivative) functions – Ninad Munshi Apr 07 '20 at 06:11

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Not quite, you must invert $\text{A}$ first and then take the derivatives of it's components - Can vectors be inverted?

This explains why:

If you have $\textbf{A}(\lambda)$ then $\textbf{A}'(\lambda)$ is essential the Jacobian $J_A$ - https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

By the inverse function theorem,

$$ (J_A)^{-1} = J_{A^{-1}}$$

You should read this page: https://en.wikipedia.org/wiki/Inverse_function_theorem