Finding y' by implicit differentiation SO HARD FOR ME help~

[r3dxP]

Find dy/dx by implicit differentiation. (Meaning find the derivative of y.. so y' = ??)

cos(x-y)=xe^x

Please help and show step by step..
the final answer should be..

y' = 1 + [(e^x)(1+x)]/[sin(x+y)]

Thanks a bunch.

 

[dextercioby]

If you know that the LHS is a function of "x" and the RHS is a function of "x" as well,then the equality A=B implies A'=B' (the equality of derivatives wrt to "x").So differentiate both sides of the eq.wrt to "x":
-[\sin(x-y)](1-\frac{dy}{dx})=e^{x}+xe^{x}\Rightarrow\frac{dy}{dx}-1=\frac{e^{x}(x+1)}{\sin(x-y)}\Rightarrow\frac{dy}{dx}=\frac{e^{x}(x+1)}{\sin(x-y)}+1
,which means that your answer was wrong.

Daniel.

 

[HallsofIvy]

It's just the chain rule. Since y is a function of x, the derivative of cos(x-y) is
-sin(x-y) times the derivative of x-y, with respect to x. Of course, the derivative of x is 1. We don't know y as a function of x so we can only write its derivative as y':
(cos(x-y))'= -sin(x-y)(1- y').

The derivative of xe^x is (by the product rule) e^x+ xe^x = e^x(1+ x) so we have

-sin(x-y)(1- y')= e^x(1+ x).

Now solve for y'.

 

[r3dxP]

Oh, Thanks. I did it so that , cos(x-y) = cosx/cosy, so i did : (-cosysinx + cosxsinyy')/(cos^2 y). Is that a valid method of doing that?

 

[HallsofIvy]

Oh, Thanks. I did it so that , cos(x-y) = cosx/cosy, so i did : (-cosysinx + cosxsinyy')/(cos^2 y). Is that a valid method of doing that?

Who were you responding to? Neither dexercioby nor I had "cos(x-y)" in our formulas.

And where did you get "cos(x-y)= cos(x)/cos(y)" from?

 

[r3dxP]

I meant when i tried solving it, i put it so that cos(x-y) is simplified to cosx/cosy, then i tried deriving that then solve the rest. I guess i did it wrong. Sorry for the confusion.

 

[dextercioby]

I meant when i tried solving it, i put it so that cos(x-y) is simplified to cosx/cosy, then i tried deriving that then solve the rest. I guess i did it wrong. Sorry for the confusion.

I hope for your sake that u are convinced that
\cos(x-y)\neq \frac{\cos x}{\cos y}
,but instead
\cos(x-y)=\cos x\cos y+\sin x\sin y

Daniel.

 

[r3dxP]

Oh thanks. I just looked up in the book and that's it! Thanks.