Question

Find the derivatives of the following:

If sin y = x sin(a + y), the prove that `("d"y)/("d"x) = (sin^2("a" + y))/sin"a"`, a ≠ nπ

Answer 1

Given sin y = x sin(a + y)   ........(1)

Differentiating with respect to x, we get

`cos  ("d"y)/("d"x) = x cos("a" + y) (0 + ("d"y)/("d"x)) + sin("a" + y) * 1`

`cos y ("d"y)/("d"x) = xcos("a" + y) ("d"y)/("d"x) + sin("a" + y)`

`cos y ("d"y)/("d"x) - xcos("a" + y) ("d"y)/("d"x) = sin("a" + y)`

`("d"y)/("d"x) = (sin("a" + y))/(cosy- xcos("a" + y))`  ........(2)

From equation (1) we have, x = `sin y/(sin("a" + y))`

Substituting for x in equation (2) we get

`("d"y)/("d"x) = (sin("a" + y))/(cosy - siny/(sin("a" + y)) * cos("a" + y))`

`("d"y)/("d"x) = (sin("a" + y))/((sin("a" + y) cosy - cos("a" + y) sin y)/(sin("a" + y))`

= `(sin^2("a" + y))/(sin ["a" + y - y])`

sin(A – B) = sinA cosB – cosA sin B

`("d"y)/("d"x) = (sin^2("a" + y))/sin "a"`