vértices ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

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∫

ddx

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

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implicit derivative dydx , y²(y²−4)=x²(x²−5)

Solution

dydx =x(2x²−5)2y(y²−2)

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Solution steps

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y²(y²−4)=x²(x²−5)

Treat y as y(x)

Differentiate both sides: 4y³dydx −8ydydx =4x³−10x

Isolate dydx : dydx =x(2x²−5)2y(y²−2)

dydx =x(2x²−5)2y(y²−2)

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

vértices ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

Solution

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