normal of y=e^{-x}*ln (x), (1, 0)

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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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normal of y=e^−xln(x), (1, 0)

Solution

y=−ex+e

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Solution steps

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Find the slope of y=e^−xln(x): −e^−xln(x)+e^−xx

Compute the slope at (1, 0): m=1e

Compute the slope of the perpendicular line: m_p=−e

Find the line with slope m=−e and passing through (1, 0): y=−ex+e

y=−ex+e

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

normal of y=e^{-x}*ln (x), (1, 0)

Solution

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