d/(dp)((5p^2-3)+(2p^2-3p^3))

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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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ddp ((5p²−3)+(2p²−3p³))

Solution

−9p²+14p

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Solution steps

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ddp ((5p²−3)+(2p²−3p³))

Apply the Sum/Difference Rule: (f±g)′=f ′±g′

=ddp (5p²)−ddp (3)+ddp (2p²)−ddp (3p³)

ddp (5p²)=10p

ddp (3)=0

ddp (2p²)=4p

ddp (3p³)=9p²

=10p−0+4p−9p²

Simplify

=−9p²+14p

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

d/(dp)((5p^2-3)+(2p^2-3p^3))

Solution

Solution steps

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