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Suppose `f(x) = tan(cos^(-1)(x))` and `g(x) = (sqrt(1-x^2))/x`.
The graph of `y = g(x)` is given.
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i. `f(x) = tan(cos^(-1)(x))`
`f^(′)(x)` | `= -1/sqrt(1-x^2) · sec^2(cos^(-1)(x))` |
`= -1/sqrt(1-x^2) · 1/(cos^2(cos^(-1)(x)))` | |
`= -1/(x^2sqrt(1-x^2))` |
`g(x) = (1-x^2)^(1/2) · x^(-1)`
`g^(′)(x)` | `= 1/2 · -2x(1-x^2)^(-1/2) · x^(-1)-(1-x^2)^(1/2) · x^(-2)` |
`= (-x)/(x sqrt(1-x^2))-sqrt(1-x^2)/(x^2)` | |
`= (-x^2-sqrt(1-x^2) sqrt(1-x^2))/(x^2 sqrt(1-x^2))` | |
`= (-x^2-(1-x^2))/(x^2sqrt(1-x^2))` | |
`= -1/(x^2sqrt(1-x^2))` | |
`=f^(′)(x)` |
ii. `f^(′)(x) = g^(′)(x)`
`=> f(x) = g(x) + c`
`text(Find)\ c:`
`f(1)` | `= tan(cos^(-1) 1)` |
`= tan 0` | |
`= 0` |
`g(1) = sqrt(1-1)/0 = 0`
`f(1) = g(1) + c`
`:. c = 0`
`:. f(x) = g(x)`