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Question
if `y = sin^(-1)[(6x-4sqrt(1-4x^2))/5]` Find `dy/dx `.
Solution
Given that
`y = sin^(-1)[(6x-4sqrt(1-4x^2))/5]`
if y = sin-1x, then `dy/dx=1/sqrt(1-x^2)`
`y = sin^(-1)[6x-4sqrt(1-4x^2)/5]`
`=> y = sin^(-1)[(6x)/5-(4sqrt(1-4x^2))/5]`
`=> y =sin^(-1)[(2x xx3)/5-(4sqrt(1-(2x)^2))/5]`
`=>y = sin^(-1)[2xx3/5-4/5sqrt(1-(2x)^2)]`
`=>y = sin^(-1)[2xxsqrt(1-(4/5)^2)-4/5sqrt(1-(2x)^2)]`
We know that
`sin^(-1)p-sin^(-1)q=sin^-1(psqrt(1-q^2)-qsqrt(1-p^2)))`
Here, p=2x and `q=4/5`
Therefore,
`y= sin^(-1)2x-sin^(-1)`
Differentiating the above function with respect to x, we have,
`dy/dx=1/sqrt(1-(2x)^2)xx2-0`
`=>dy/dx=2/sqrt(1-4x^2)`
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