Question

differentiate y = 8e^x + 2/3^squarerootx

Answer 1

first part is easy enough since
\[\frac{d}{dx}e^x=e^x\]

Answer 2

so the first part is zero right?

Answer 3

second part use
\[\frac{d}{dx}[b^x]=b^x\ln(b)\]so by the chain rule
\[\frac{d}{dx}b^{f(x)}=\ln(b)\times b^{f(x)}\times f'(x)\]

Answer 4

oh heck no

Answer 5

the derivative of \(8e^x\) is not \(0\) since \(8e^x\) is not a constant

Answer 6

the derivative of \(e^x\) is \(e^x\)
leave the \(8\) in front

Answer 7

what do i do about