Question

Ned someone to check my Calculus work.

Answer 1

Am I correct?

Answer 2

@thomaster @TuringTest

Answer 3

i like it up to the 4th or 5th part .. after that its just algebraing it to death :)

Answer 4

Do you know if I was correct? It's the only one I am questioning.

Answer 5

\[y=a~exp(x^3)\]
\[ln(y)=x^3~ln(a)\]
\[y'/y=b^3~a'/a+3x^2ln(a)\]
\[y'=aexp(x^3)~b^3~a'/a+aexp(x^3)~3x^2ln(a)\]
\[y'=aexp(x^3-1)~b^3~a'+aexp(x^3)~3x^2ln(a)\]
if im keepin gtrack of it correctly

Answer 6

pfft, i change b to x halfway thru and forgot a b :)

Answer 7

As an alternative approach you might want to write
\[\Large y= (\sin x)^{x^3} \] as an exponential property using the \(\exp\) function:
\[\large y= \exp ( x^3 \log (\sin x)) \] which is basically the same method as you've choose above, but is a bit more handy in notation.

Answer 8

Alright, thank you @Spacelimbus and @guest.100 :)

Answer 9

So, I was right?

Answer 10

whats the wolf say? :)

Answer 11

Haha, I tried to use it, that's site is all jacked up.

Answer 12

that.*

Answer 13

x^2 (sin x) ^(x^3-1) (x cos(x) + 3 sin(x) ln(sin(x)) )

Answer 14

\[y'=a^{x^3-1}~x^3~a'+a^{x^3}~3x^2ln(a)\]
\[y'=x^2~a^{x^3-1}[x~a'+3a~ln(a)]\]
i think it likes mine :)

Answer 15

yeah, factor an x^2 out of yours and you match