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find dy/dx from y = 2x^3 * ln(ax)
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yayy another product ruleeee
What part you getting stuck on? :O
\[(fg)'=f'g+fg'\]
with \[f(x)=2x^3,f'(x)=6x^2,g(x)=\ln(ax), g'(x)=\frac{1}{x}\]
@ zepdrix, i just need to check my answers lol... i keep thinking that i did things wrong
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Oh i see :)
isn't it chain rule with ln(ax)?
i meant product rule?
chain rule? :o oh yes i suppose it is. \[[\ln(ax)]'=\frac{ 1 }{ ax }(ax)'=\frac{ 1 }{ ax }a=\frac{ 1 }{ x }\] But you still get what satalite posted :3 after simplifying.
Well hopefully they intended for A to be a constant, which is why we wouldn't apply the product rule :O it's not a variable thingy
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well there's no where that says a is a constant :(
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