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प्रश्न
`log (x + sqrt(x^2 + "a"))`
उत्तर
Let y = `log (x + sqrt(x^2 + "a"))`
Differentiating both sides w.r.t. x
`"dy"/"dx" = "d"/"dx" log (x + sqrt(x^2 + "a"))`
= `1/(x + sqrt(x^2 + "a")) * "d"/"dx" (x + sqrt(x^2 + "a"))`
= `1/(x + sqrt(x^2 + "a")) * [1 + 1/(2sqrt(x^2 + "a")) xx "d"/"dx" (x^2 + "a")]`
= `1/(x + sqrt(x^2 + "a")) * [1 + 1/(2sqrt(x^2 + "a")) * 2x]`
= `1/(x + sqrt(x^2 + "a")) * [1 + x/(sqrt(x^2 + "a"))]`
= `1/(x + sqrt(x^2 + "a")) * ((sqrt(x^2 + "a") + x)/(sqrt(x^2 + "a")))`
= `1/(sqrt(x^2 + "a")`
Hence. `"dy"/"dx" = 1/sqrt(x^2 + "a")`
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