I\'m looking for step by step details on how to take the partial derivative with

Question

I'm looking for step by step details on how to take the partial derivative with respect to 'Y' then respect to "X" of sin (x+y).

Short cut not allowed so I will need to also use the sum/difference identitfies plus show the step by step details using the derivative rule/formula below

sum/difference

Sin (x+Y) = [sin(x)cos(y)+ cos(x)sin(y)]

Cos(x+Y) = [cos(x)cos(y)+ sin(x)sin(y)]

derivative rule/formula

[f(x+?x)-f(x)/?X]

.

Please help by showing step by ste details with respec to Y then resect to x.  

Thanks

Explanation / Answer

Let x,y in R^2 and let h>0

df/dx(x,y) = lim(h->0) (f(x+h,y)-f(x,y))/h

df/dy(x,y) = lim(h->0) (f(x,y+h)-f(x,y))/h

(sin(x+h+y)-sin(x+y))/h = (sin(x+y)cos(h)+cos(x+y)sin(h)-sin(x+y))/h = sin(x+y)(cos(h)-1)/h + cos(x+y)sin(h)/h

Now remember that :

lim(h->0) sin(h)/h = 1

lim(h->0) (cos(h)-1)/h = lim(h->0)-sin(h)/1=0 (L'hospital rule)

Hence lim(h->0)(sin(x+h+y)-sin(x+y))/h= cos(x+y)

So df/dx = cos(x+y)

Since (f(x+h,y)-f(x,y))/h=(sin(x+y+h)-sin(x+y))/h = (sin(x+h+y)-sin(x+y))/h=(f(x,y+h)-f(x,y))/h

Then also df/dy = cos(x+y)

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