Question

h. find y’ and y” for y = 9e^((〖3x〗^2+4x-5))

Answer 1

9e\[y = 9e ^{3x ^{2}}+\]

Answer 2

\[y = 9e ^{(3x ^{2} + 4x - 5}\]

Answer 3

y=9e^(g(x))
y'=9g'(x)e^(g(x))
y''=9[g''(x)e^(g(x))+g'(x)*g'(x)*e^(g(x))]

Answer 4

was going to ask if you can set up few more for me

Answer 5

ok

Answer 6

i. y = ln abs value(8x^3 + 3^8x)

Answer 7

y=ln|g(x)|
y'=(g'(x))/(g(x))

Answer 8

and if you have y=a^x
lny=lna^x
lny=xlna
y'/y=lna
y'=ylna
y'=(a^x)lna

so if you have y=a^x, then y'=(a^x)(lna)

Answer 9

y = logbase7((5x^3 -2x^2 + 1)/ (3x^4-5))

Answer 10

where a is a constant bigger than 0

Answer 11

thnx

Answer 12

do you want me to start a new thread so you can get medals.

Answer 13

\[y=\log_7(g(x))=\frac{\ln(g(x))}{\ln7}\]
it doesn't seem like equation editor is working that sucks

y=logbase7(g(x))=[ln(g(x))]/[ln(7)]
so
y'=1/(ln7)* g'(x)/g(x)

Answer 14

lol

Answer 15

go ahead and make new post just in case i have to leave
k?
i hope you can understand the formulas i'm giving you

Answer 16

trying to really hard ha

Answer 17

not a math guru... =)

Answer 18

you can do it!
:)