Find dy/dx xcos(y)+ycos(x)=1

Problem

x*cos(y)+y*cos(x)=1

Solution

1. Differentiate both sides with respect to x using implicit differentiation.

d()/d(x)*(x*cos(y)+y*cos(x))=d()/d(x)*(1)

2. Apply the product rule to both terms on the left side, remembering that y is a function of x

(d(x)*cos(y))/d(x)+(d(y)*cos(x))/d(x)=0

3. Expand the derivatives using the product rule and the chain rule for terms involving y

(1)*cos(y)+x*(−sin(y))d(y)/d(x)+d(y)/d(x)*cos(x)+y*(−sin(x))=0

4. Simplify the expression by removing parentheses and organizing the terms.

cos(y)−x*sin(y)d(y)/d(x)+cos(x)d(y)/d(x)−y*sin(x)=0

5. Isolate the terms containing d(y)/d(x) on one side of the equation.

cos(x)d(y)/d(x)−x*sin(y)d(y)/d(x)=y*sin(x)−cos(y)

6. Factor out d(y)/d(x) from the left side.

d(y)/d(x)*(cos(x)−x*sin(y))=y*sin(x)−cos(y)

7. Solve for d(y)/d(x) by dividing both sides by the factor in parentheses.

d(y)/d(x)=(y*sin(x)−cos(y))/(cos(x)−x*sin(y))

Final Answer

d(y)/d(x)=(y*sin(x)−cos(y))/(cos(x)−x*sin(y))

Want more problems? Check here!