Why this image is true?

The image's correctness arises from the formal rules of vector calculus and automatic differentiation applied to scalar functions of vector variables. Let's dissect why each computational step in the image is valid.

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1. Vector Definition and Scalar FunctionThe image first sets

$$A = <\text{span} \text{class}='\text{cite} \text{cite}-1'>[1]</\text{span}><\text{span} \text{class}='\text{cite} \text{cite}-2'>[2]</\text{span}>$$

which defines a two-dimensional vector $A$ with $A<\text{span} \text{class}='\text{cite} \text{cite}-3'>[3]</\text{span}> = 5$ and $A<\text{span} \text{class}='\text{cite} \text{cite}-4'>[4]</\text{span}> = 7$. This representation is consistent with standard vector notation in mathematics and engineering, ensuring that subsequent elementwise operations are meaningful and unambiguous.**2. Calculating the Scalar Function $L$**Next, the scalar $L$ is computed as

$$L = 3A<\text{span} \text{class}='\text{cite} \text{cite}-3'>[3]</\text{span}>A<\text{span} \text{class}='\text{cite} \text{cite}-4'>[4]</\text{span}>$$

Substituting the components of $A$ yields:

$$L = 3 \times 5 \times 7 = 105$$

This step reflects basic algebraic operations and scalar multiplication, fundamental to both linear algebra and elementary calculus.3. Differentiation with Respect to a VectorThe core operation is calculating the derivative of $L$ with respect to the vector $A$:

$$\frac{dL}{dA}$$

Since $L$ is a function of both $A<\text{span} \text{class}='\text{cite} \text{cite}-3'>[3]</\text{span}>$ and $A<\text{span} \text{class}='\text{cite} \text{cite}-4'>[4]</\text{span}>$, the gradient (vector of partial derivatives) is:

• $$$$

\frac{\partial L}{\partial A<span class='cite cite-3'>[3]</span>} = 3A<span class='cite cite-4'>[4]</span> = 21

- $$ \frac{\partial L}{\partial A<\text{span} \text{class}='\text{cite} \text{cite}-4'>[4]</\text{span}>} = 3A<\text{span} \text{class}='\text{cite} \text{cite}-3'>[3]</\text{span}> = 15

Therefore,

$$\nabla_A L = \left[21, 15\right]$$

This is precisely the gradient vector, which is the standard output for the differential of a scalar field with respect to a vector in $\mathbb{R}^n$ [4][5].4. Summation of Gradient ComponentsThe final line in the image presents $= 36$, which is the sum of the gradient components:

$$21 + 15 = 36$$

This step is not part of traditional calculus, but it is frequently encountered in computational contexts—particularly in automatic differentiation (AD) systems—where summing the components of the gradient is required for optimization or further computation, such as when calculating the total sensitivity of $L$ with respect to all elements of $A$ [4][6][7].

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Supporting Theoretical Context

• The vector differentiation, as shown, adheres to classical rules of vector calculus and is compatible with modern automatic differentiation techniques. These techniques evaluate gradients efficiently and exactly for any smooth composite function, as occurs here [4][6][7].
• The methodology used is well-founded in both theoretical mathematics and computational practice, as discussed in depth in studies on the implementation of automatic differentiation for vector-valued functions and their gradients [4][6][8].
• The summation step reflects a pragmatic approach in algorithmic differentiation, often used when a scalar sensitivity measure is needed, or when projecting the gradient in a direction of interest [4][6].

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Conclusion

Each step in the image follows rigorously from established mathematical principles in vector calculus and is mirrored in the formal structure of automatic differentiation. The numerical and algebraic outcomes are correct, and the presentation (including the summing of partial derivatives) aligns with real-world computational workflows [4][5][6][7].

Thus, the image is true because it exactly implements the gradient computation pipeline, as supported by decades of mathematical research and computational science development.