Quick question (1 Viewer)

[<Stretch>]

Could you please show me how to differentiate this with all working:

y = e xy (cosx)

Thanks in advance

 

[pLuvia]

Is that suppose to be y=e^xycosx?

or y=e^xxy(cosx)

 

[<Stretch>]

Sorry, should have been clearer. Its y=e^cosx

 

[pLuvia]

Ok that makes more sense

Oh you wanted working lol, ok here

well the general formula is y'=f'(x)e^f(x)

.:f(x)=cosx
f'(x)=-sinx

And e^cosx=e^cosx

And so you get this answer

y'=-sinx*e^cosx

 

[<Stretch>]

Thanks for the speedy reply. Your a lifesaver. But is there any working you can show me to link the two?

Sorry working just noticed. Champion.

 

[Riviet]

Basically you get the original function multiplied by the derivative of the function in the index.

 

[pLuvia]

Oh I edited the post, when I read you wanting working

the general formula is y'=f'(x)e^f(x), and you when you d (e^f(x))/dx = e^f(x)

 

[Mountain.Dew]

to be REALLY general, that was merely an application of the chain rule.

y = e^cosx

so, using dy/dx = (dy/du) * (du/dx), it is:

dy/dx = [d(e^cosx)/d(cosx)] * [d(cosx) / dx]
= e^cosx*[-sinx]
=-sinx*e^cosx