Find dy/dx sin(x^2y^2)=x

Problem

sin(x^2*y^2)=x

Solution

1. Differentiate both sides with respect to x using the chain rule on the left side.

d(sin(x^2*y^2))/d(x)=d(x)/d(x)

2. Apply the chain rule to the sine function, which involves the derivative of the inner expression x^2*y^2

cos(x^2*y^2)⋅(d(x^2)*y^2)/d(x)=1

3. Apply the product rule to differentiate x^2*y^2 treating y as a function of x

cos(x^2*y^2)⋅(2*x*y^2+x^2⋅2*yd(y)/d(x))=1

4. Distribute the cosine term into the parentheses to isolate the terms containing d(y)/d(x)

2*x*y^2*cos(x^2*y^2)+2*x^2*y*cos(x^2*y^2)d(y)/d(x)=1

5. Isolate the term with d(y)/d(x) by subtracting 2*x*y^2*cos(x^2*y^2) from both sides.

2*x^2*y*cos(x^2*y^2)d(y)/d(x)=1−2*x*y^2*cos(x^2*y^2)

6. Solve for dy/dx by dividing both sides by the coefficient 2*x^2*y*cos(x^2*y^2)

d(y)/d(x)=(1−2*x*y^2*cos(x^2*y^2))/(2*x^2*y*cos(x^2*y^2))

Final Answer

d(y)/d(x)=(1−2*x*y^2*cos(x^2*y^2))/(2*x^2*y*cos(x^2*y^2))

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