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

Solutions > Power Mod Calculator >

interceptos (x^2+2x-1)*(2x^2-3x+6)

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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_□

π

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∫

ddx

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

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Steps Examples

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ddx (sin⁷(x))

Solution

7sin⁶(x)cos(x)

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Solution steps

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Second Partial Derivative Test definition

Suppose that (x, y)=(a, b) is a critical point of f(x, y)
and D(x, y)=∂²f∂ x² ∂²f∂ y² −(∂²f∂ x∂ y )². Then,
If D(a, b) > 0 and ∂²f∂ x² <0 then the point (a, b) is a local maximum.
If D(a, b) > 0 and ∂²f∂ x² >0 then the point (a, b) is a local minimum.
If D(a, b)<0 then the point (a, b) is a saddle point.
If D(a, b)=0 then test failed and the point (a, b) can be a local maximum, local minimum, saddle or neither.

Encontrar os pontos críticos: (0, 0)

∂²f∂ x² =2

D(x, y)=4

Verificar o sinal de em cada ponto crítico

Verificar o ponto crítico (0, 0): Mínimo

Mínimo(0, 0)

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