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प्रश्न
If `y=sin^-1(3x)+sec^-1(1/(3x)), ` find dy/dx
योग
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उत्तर
`y=sin^-1(3x)+sec^-1(1/(3x))`
`dy/dx=d/dxsin^-1(3x)+d/dxsec^-1(1/(3x))`
`dy/dx=3/sqrt(1 -(3x)^2) + ((-1)/(3x^2))/(1/(3x)sqrt((1/(3x)^2-1)))`
`dy/dx= 3/sqrt(1-9x^2) - 1/("X" sqrt((1-9x^2))/(3|x|))`
`= 3/sqrt(1-9x^2) - (3|x|)/("X" sqrt(1-9x^2))`
`= 3/(sqrt (1 -9x^2)) (1 - |X|/X)`
= 0 X > 0
= `6/sqrt(1-9"x"^2)` X < 0
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