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What is the derivative of 4^x lnx+x tanx+4e^x
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\[\frac{ d }{ dx }a^x = a^xlog(a)\] \[\frac{ d }{ dx }ln(x) = \frac{1}{x}\]
would the + make the equation f+g+h?
that's correct! But in the question, you ALSO have f(g)+h(i) + k so you have to use product rule
so my answer would be f1(g)f(g1)+h1(i)h(i1)+k1? with 1 meaning prime
Yes, however, prime is written like this: f' just to make it less confusing. But you got it down :)
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you sir, are a genius and deserve a bueno!
just as a final confirmation my answer was (4^x log4 lnx)(4^x 1/x)+(tanx)(xsec^2x)+4e^x
Looks right to me.
thanks alot
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