The Student Room Group
Reply 1
Have you tried using the quotient rule?
Reply 2
let u= sin3x v=e^x

du/dx= 3cos3x
dv/dx = e^x

f'(x)= [ v(du/dx) - u(dv/dx) ] / v^2
= [ e^x(3cos3x ) - sin3x (e^x) ] / (e^x)^2
=[ e^x ( 3cos3x-sin3x) ] / (e^x)^2
= ( 3cos3x-sin3x) ] / (e^x).
Reply 3
FZka
let u= sin3x v=e^x

du/dx= 3cos3x
dv/dx = e^x

f'(x)= [ v(du/dx) - u(dv/dx) ] / v^2
= [ e^x(3cos3x ) - sin3x (e^x) ] / (e^x)^2
=[ e^x ( 3cos3x-sin3x) ] / (e^x)^2
= ( 3cos3x-sin3x) ] / (e^x).



aaah i seee thanks:biggrin: and one more thing
is (e^x)^2=e^2x or e^x^2
Reply 4
e^2x
Reply 5
DLS
e^2x


thank You for clearing that up:biggrin: now im all done! just need to revise solving equation chapter which won't take long:smile: then off to bed and wake up early 8am and start refreshing my mind lol
Reply 6
Remarqable M
aaah i seee thanks:biggrin: and one more thing
is (e^x)^2=e^2x or e^x^2

(x^n)^m = x^nm
Reply 7
wat wud happen in the situation (3x^2)^3 ???
Reply 8
Knight91
wat wud happen in the situation (3x^2)^3 ???


27x^6
Reply 9
Remarqable M
how to differentiate this:

f(x)=sin3x/e^x

thnx
note: rep+ available the day after exam.


:innocent:

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