неявная производная (dy)/(dx),x^2(x-y)^2=x^2-y^2

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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prove sin(2x)sin(x) −cos(2x)cos(x) =sec(x)

Solution

True

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Solution steps

One step at a time

(

4	4	2	3	−2
0	1	−2	−2	2
6	12	11	2	−4
9	20	10	10	−6
15	28	14	5	−3

)

Matrix eigenvalues

Let A be a square matrix, η a vector and λ a scalar that satisfy Aη=λη,
then λ is called the eigenvalue associated with the eigenvector η of A

The eigenvalues of A are the roots of the characteristic equation det(A−λ I)=0

det((

4	4	2	3	−2
0	1	−2	−2	2
6	12	11	2	−4
9	20	10	10	−6
15	28	14	5	−3

)−λ(

1	0	0	0	0
0	1	0	0	0
0	0	1	0	0
0	0	0	1	0
0	0	0	0	1

)): −λ⁵+23λ⁴−206λ³+898λ²−1905λ+1575

Solve −λ⁵+23λ⁴−206λ³+898λ²−1905λ+1575=0: λ=3 with multiplicity of 2, λ=5 with multiplicity of 2, λ=7

The eigenvalues are:

λ=3 with multiplicity of 2, λ=5 with multiplicity of 2, λ=7

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

неявная производная (dy)/(dx),x^2(x-y)^2=x^2-y^2

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