exponential differentiation (1 Viewer)

[C6H12O6-Glucose]

can somebody please show me how to find y' for the following:

e<SUP>x<SUP>4</SUP>y</SUP> = x + y

ty

 

[GUSSSSSSSSSSSSS]

e^(y.x^4) = x + y
differentiate implicitly:

(y.4x^3 + dy/dx . x^4)e^(y.x^4) = 1 + dy/dx

re-arranging

dy/dx . (x^4 . e^(y.x^4)) - dy/dx = 1 - 4yx^3 . e^(y.x^4)
dy/dx = (1 - 4yx^3 . e^ (y.x^4)) / (x^4 . e^(y.x^4))

is that rite??? its too hard to see any simplifying on the computer screen lol xD

 

[bored of sc]

can somebody please show me how to find y' for the following:

e<SUP>x<SUP>4</SUP>y</SUP> = x + y

ty

Implicit differentiation?

To find y' or dy/dx you differentiate each term separately and wherever you have differentiated y you multiply it by dy/dx.

e^x4y = x + y

Differentiating:

4x³e^x4y + e^x4y.dy/dx = 1 + dy/dx

Now replace e<SUP>x<SUP>4</SUP>y</SUP> with x + y

4x³(x+y) + (x+y)dy/dx = 1 + dy/dx

(x+y)dy/dx -dy/dx = 1 - 4x³(x+y)
(x+y-1)dy/dx = ditto
dy/dx = [1-4x³(x+y)] / (x+y-1)

 

[GUSSSSSSSSSSSSS]

lol we got different answers ....i cbf checking mine xD

 

[bored of sc]

lol we got different answers ....i cbf checking mine xD

I think you'd be right. Actually our answers are pretty similar.

 

[Drongoski]

Is my answer wrong ??

Edit

On reflection, much easier to implicitly differentiate direct without all the above hassle.

 

[undalay]

duuuuuuude beta glucose > alpha

 

[Drongoski]

e^(y.x^4) = x + y
differentiate implicitly:

(y.4x^3 + dy/dx . x^4)e^(y.x^4) = 1 + dy/dx

re-arranging

dy/dx . (x^4 . e^(y.x^4)) - dy/dx = 1 - 4yx^3 . e^(y.x^4)
dy/dx = (1 - 4yx^3 . e^ (y.x^4)) / (x^4 . e^(y.x^4))

is that rite??? its too hard to see any simplifying on the computer screen lol xD

GUS ..SSS

Don't forget: e^(y.x^4) = x+y

 

[GUSSSSSSSSSSSSS]

GUS ..SSS

Don't forget: e^(y.x^4) = x+y

LOL drongoski i just cbb simplifying xD

 

[GUSSSSSSSSSSSSS]

LOL drongoski i just cbb simplifying xD

and NICE way of writing my name Drongoski xD
once again you make my day !!!!!! =]]]