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Question
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `cos^-1((7x^4 + 5y^4)/(7x^4 - 5y^4)) = tan^-1a`
Solution
`cos^-1((7x^4 + 5y^4)/(7x^4 - 5y^4)) = tan^-1a`
`(7x^4 + 5y^4)/(7x^4 - 5y^4)` = cos(tan–1 a) = b
`(7x^4 + 5y^4)/(7x^4 - 5y^4)` = b
7x4 + 5y4 = b(7x4 – 5y4)
7x4 + 5y4 = 7bx4 – 5by4
5y4 + 5by4 = 7bx4 – 7x4
5y4(1 + b) = 7x4(b –1)
`(5y^4)/(7x^4) = (b - 1)/(1 + b)`
`y^4/x^4 = (7(b - 1))/(5( 1 + b)` = x
`y^4/x^4` = c....(1)
y4 = cx4
Differentiating both sides w.r.t. x, we get
`4.y^3"dy"/"dx"` = c.4x3
`"dy"/"dx" = (c.4x^3)/(4y^3)`
`"dy"/"dx" = (c.x^3)/y^3`
`"dy"/"dx" = y^4/x^4.x^3/y^3` ...from..(1)
`"dy"/"dx" = y/x`.
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