Find dy/dx by implicit differentiation for y cos x = 1 + sin(xy)

Question

Find dy/dx by implicit differentiation for y cos x = 1 + sin(xy)

Answer 1

First, differentiate each side of the equation with respect to x. This means that whenever we differentiate y, we need to multiply the result by dy/dx. Let's do the left side first:

 

We need to use the Product Rule-- uv' + vu'. Let u=y and v=cos x.

Then,  d/dx (y cos x) = y (- sin x) + cos x dy/dx

 

Simplified, the left side should read:    -y sin x + cos x dy/dx

 

Now for the right side:

 

The derivative of 1 is 0, so we eliminate that part. You need to use the Chain Rule to find the derivative of sin(xy), and the Product Rule again to get the derivative of xy.

 

d/dx (sin (xy))= (cos(xy)) (x dy/dx + y)

 

Simplified right side:   x cos(xy) dy/dx + y cos(xy)

 

Now let's rewrite our equation:

 

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

 

Now all we need to do is solve the equation for dy/dx.

First, add and subtract to get all terms with dy/dx on the same side.

 

This gives

 

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

 

Factor both sides.

 

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

 

Now we can divide to get dy/dx alone.

 

dy/dx =  [y(cos(xy) + sin x)] / [cos x - x cos(xy)]

 

And that's it. I hope this helps, and that the punctuation wasn't too confusing.

 

Samantha B.

Answer 2

The following is only supplementary and may be of interest:

 

It may be of interest that this problem could have been done in an alternative (somewhat shorter) manner.  However, if you are confused by the following, do not worry; it is technically derived from the chain rule (for partial derivatives---a multivariable calculus topic).  However, I think this may be interesting FYI:

 

Given f(x,y)=c (for some constant c), then dy/dx= -fx/fy, where fx is the partial derivative of f wrt x and fy is the partial wrt y.

 

Here, we can manipulate the equation : ycos(x)-sin(xy)=1 to have the above form so that the Left Hand Side is f(x,y) and the RHS is our "c."

 

fx=-ysinx-ycos(xy), fy=cosx-xcos(xy)

 

Hence y'= y(sinx+cos(xy))/(cosx-xcos(xy))