How do you differentiate #sinx+cosy=sinxcosy#?

Answer 1

Scarlett Allison

I found:
#(dy)/(dx)=(cos(x)(cos(y)-1))/(sin(y)(sin(x)-1)#

You need to remember the #y# will be a function of #x# and differentiate accordingly, so you have:

#cos(x)-sin(y)(dy)/(dx)=cos(x)cos(y)-sin(x)sin(y)(dy)/(dx)#

where on the right I used the Product Rule;

now collect #(dy)/(dx)#:

#(dy)/(dx)=(-cos(x)+ cos(x)cos(y))/(-sin(y)+sin(x)sin(y))=(cos(x)(cos(y)-1))/(sin(y)(sin(x)-1)#

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Answer 2

Samantha Adkison

To differentiate the equation sin(x) + cos(y) = sin(x)cos(y) with respect to x, we use implicit differentiation.

Differentiating both sides of the equation with respect to x, we get:

cos(x) - sin(y) * (dy/dx) = cos(x) * cos(y) - sin(x) * sin(y) * (dy/dx)

Rearranging terms to solve for (dy/dx), we get:

dy/dx = (cos(x) - cos(x) * cos(y)) / (sin(y) * sin(x) - cos(y))

This is the derivative of y with respect to x.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.