d/(da)((11a^{15})(a^4)(a^{-3}))

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

☐²

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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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dda ((11a¹⁵)(a⁴)(a⁻³))

Solution

176a¹⁵

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dda ((11a¹⁵)(a⁴)(a⁻³))

Take the constant out: (a·f)′=a·f ′

=11dda (a¹⁵a⁴a⁻³)

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

f=a¹⁵, g=a⁴a⁻³

=11(dda (a¹⁵)a⁴a⁻³+dda (a⁴a⁻³)a¹⁵)

dda (a¹⁵)=15a¹⁴

dda (a⁴a⁻³)=1

=11(15a¹⁴a⁴a⁻³+1· a¹⁵)

Simplify 11(15a¹⁴a⁴a⁻³+1· a¹⁵): 176a¹⁵

=176a¹⁵

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

d/(da)((11a^{15})(a^4)(a^{-3}))

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