LyleX^Y
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- 0
can you guyz help me with a quick derivative?
eeeex
eeeex
LyleX^Y said:awww sorry this isn't a homework actually I am just thinking of a problem my teacher would do. hehe sorry
i got
d/dx=(eeeex)*(d/dx eeex)*(d/dx eex)*(d/dx ex)
i just can't get past that point i don't know how to do the product rule with 3 values.
gabbagabbahey said:Well, so far so good(although it's the chain rule you're using, not the product rule)
Now try computing each of the derivatives in that product (use the chain rule again)... what are (d/dx eeex), (d/dx eex) and (d/dx ex)?
LyleX^Y said:dont i have to use the product rule for the 3 that i haven't take the d/dx of?
"(d/dx eeex), (d/dx eex) and (d/dx ex)?
LyleX^Y said:awww sorry this isn't a homework actually I am just thinking of a problem my teacher would do. hehe sorry
i got
d/dx=(eeeex)*(d/dx eeex)*(d/dx eex)*(d/dx ex)
i just can't get past that point i don't know how to do the product rule with 3 values.
cragar said:could we differentiate it like a^x by so it owuld be lna*a^x