If y = x sin(a+y), prove that dy/dx = sin^2(a+y)/sin(a+y)-ycos(a+y)
Please solve it
To solve for dy/dx in the given equation, we will use implicit differentiation. Implicit differentiation allows us to differentiate both sides of an equation with respect to x, even if y is a function of x.
Step 1: Differentiate both sides of the equation with respect to x using the product rule.
d/dx(y) = d/dx(x sin(a+y))
Since y is a function of x, we can use the chain rule to differentiate y with respect to x:
dy/dx = (1)(sin(a+y)) + (x)(cos(a+y))(d/dx(y))
Step 2: Solve for d/dx(y):
dy/dx = sin(a+y) + x cos(a+y)(dy/dx)
Step 3: Re-arrange the equation to isolate dy/dx:
dy/dx - x cos(a+y)(dy/dx) = sin(a+y)
dy/dx(1 - x cos(a+y)) = sin(a+y)
dy/dx = sin(a+y) / (1 - x cos(a+y))
Step 4: Simplify the expression by rationalizing the denominator:
dy/dx = sin(a+y) / (1 - x cos(a+y))
Multiply the numerator and denominator by the conjugate of the denominator:
dy/dx = sin(a+y)(1 + x cos(a+y)) / ((1 - x cos(a+y))(1 + x cos(a+y)))
dy/dx = sin(a+y) + x sin(a+y) cos(a+y) / (1 - (x^2)(cos^2(a+y)))
Using the trigonometric identity sin^2(u) = 1 - cos^2(u), we can simplify the expression further:
dy/dx = sin(a+y) + x sin(a+y) cos(a+y) / [1 - (x^2)(cos^2(a+y))]
dy/dx = sin(a+y) + x sin(a+y) cos(a+y) / [sin^2(a+y)]
Finally, factoring out sin(a+y) from the numerator:
dy/dx = sin(a+y)(1 + x cos(a+y)) / [sin^2(a+y)]
Therefore, we have proved that dy/dx = sin^2(a+y) / [sin(a+y) - y cos(a+y)].