Question

Find the derivative of (sin x)^1/x.

Answer 1

That's the function you want to derivate??\[(\frac{ \sin x }{ x})'\]

Answer 2

the equation is\[(\sin x)^\frac{ 1 }{ x }\]

Answer 3

chain rule
let
u = sin x ..... du = cos x
\[\frac{d}{dx} u^{1/x} = \frac{1}{x}u^{1/x - 1} * du\]

Answer 4

wait i think i messed that up :{

Answer 5

so would the answer be: \[\frac{ 1 }{ x }(\sin x)^{\frac{ 1 }{ x }-1}\cos x\] ?

Answer 6

yeah but i dont think i did derivative right for u^1/x

Answer 7

yeah, that's the part that fried my brain :/ not sure how to approach the 1/x

Answer 8

ok i got it, its just another level of chain rule
let
w = 1/x ..... dw = -1/x^2
\[\frac{d}{dx} u^w = \ln (u) u^w * du * dw\]

Answer 9

ahhh.. I see it! Thank you so much!!

Answer 10

your welcome

Answer 11

where did the ln(u) come from?

Answer 12

derivative of an exponential function
\[\frac{d}{dx} a^x = \ln(a) a^x\]
thats why derivative of e^x is just e^x

Answer 13

yes, of course

Answer 14

that was well thought out.
who comes up with these problems?

Answer 15

Monash University :( I have a maths assignment.. My heart broke when I saw the questions

Answer 16

ah! that explains it. They dont make things easy

Answer 17

not at all