implizite ableitung (dy)/(dx),y2=7x^2+12x

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

☐²

x^☐

√☐

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log_□

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∫

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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Steps Examples

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diagonalize (

−4	−17
2	2

)

Solution

P=(

−3+5`i`	−3−5`i`
2	2

), D=(

−1+5`i`	0
0	−1−5`i`

), P⁻¹=(

−`i`10	14 −`i`320
`i`10	14 +`i`320

)

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Solution steps

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(

−4	−17
2	2

)

Matrix diagonalization

A matrix A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such that A=PDP⁻¹

Find the eigenvalues for (

−4	−17
2	2

): λ=−1+5i, λ=−1−5i

The diagonal matrix D is composed of the Eigenvalues:

D=(

−1+5`i`	0
0	−1−5`i`

)

Find the eigenvectors for (

−4	−17
2	2

): (

−3+5i

), (

−3−5i

)

The Eigenvectors corresponding to the Eigenvalues in D compose the columns of P :

P=(

−3+5`i`	−3−5`i`
2	2

)

Find P⁻¹: (

−`i`10	14 −`i`320
`i`10	14 +`i`320

)

Verify that A=PDP⁻¹

Solve PDP⁻¹: (

−4	−17
2	2

)

P=(

−3+5`i`	−3−5`i`
2	2

), D=(

−1+5`i`	0
0	−1−5`i`

), P⁻¹=(

−`i`10	14 −`i`320
`i`10	14 +`i`320

)

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−4	−17
2	2

) or choose below

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

implizite ableitung (dy)/(dx),y2=7x^2+12x

Solution

Solution steps

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