derivada implícita (d^2y)/(dx^2), e^x+e^y=1

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

☐²

x^☐

√☐

^□√☐

□□

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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normal of f(x)=x⁴+2e^x, (0, 2)

Solution

f(x)=−12 x+2

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ddx ((2x−5)e^x)

Apply the Product Rule: (f·g)′=f ′·g+f·g′

f=2x−5, g=e^x

=ddx (2x−5)e^x+ddx (e^x)(2x−5)

ddx (2x−5)=2

ddx (e^x)=e^x

=2e^x+e^x(2x−5)

Simplify 2e^x+e^x(2x−5): 2e^xx−3e^x

=2e^xx−3e^x

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

derivada implícita (d^2y)/(dx^2), e^x+e^y=1

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