Find dy/dx e^ycos(x)=1+sin(xy)

Problem

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

Solution

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

2. Apply the product rule to the left side, d()/d(x)*[ey*cos(x)]

d(ey)/d(x)*cos(x)+eyd(cos(x))/d(x)

eyd(y)/d(x)*cos(x)−ey*sin(x)

3. Apply the chain rule and product rule to the right side, d()/d(x)*[1+sin(x*y)]

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

cos(x*y)*(y+xd(y)/d(x))

4. Equate the derivatives from both sides.

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

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

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

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

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

7. Solve for the derivative by dividing both sides by the expression in parentheses.

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

Final Answer

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

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