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`cos ^(-1) x^(2)` के सापेक्ष` tan ^(-1) {(sqrt(1+x^(2))sqrt(-1-x^(2)))/(sqrt(1+x^(2))sqrt(1-x^(2)))}` का अवकलन कीजिए| |
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Answer» माना `y_1 =tan ^(-1){(sqrt(1+x^(2))sqrt(-x^(2)))/(sqrt(1+x^(2))sqrt(1-x^(2)))}` व ` " "y_2 =cos ^(-1)x ^(2)` ` rArr " "x^(2) =cos y^(2)` `x^(2)` का मान समीकरण (1 ) में रखने पर ` y_1 =tan ^(-1) {(sqrt(1+cos y_2)-sqrt(1-cos y_2))/(sqrt(1+cos y_2 ) +sqrt(1-cos y_2))}` ` " "=tan ^(-1) {(sqrt(2cos ^(2) (y_2//2))-sqrt(2sin ^(2) (y_2//2)))/(sqrt(2cos ^(2)(y_2//2))+sqrt(2sin (y_2//2)))}` ` " "=tan ^(-1) ""{(cos (y_2//2)-sin (y_2//2))/(cos (y_2//2)+sin (y_2//2))}` ` (cos y_2//2) ` से भाग करने पर, ` " "= tan ^(-1) {(1- tan (y_2//2))/(1+tan (y_2//2))}` ` " "= tan ^(-1) {tan ((pi)/(4) -(y_2)/(2))}` ` " "=((pi)/(4)-(y_2)/(2))rArr " "(dy_1)/(dy_2)=(-1)/(2)` |
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