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Question
Find `"dy"/"dx"`, if : x = `sqrt(a^2 + m^2), y = log(a^2 + m^2)`
Solution
x = `sqrt(a^2 + m^2), y = log(a^2 + m^2)`
Differentiating x and y w.r.t. x, we get
`"dx"/"dm" = "d"/"dm"(sqrt(a^2 + m^2))`
= `(1)/(2sqrt(a^2 + m^2))."d"/"dm"(a^2 + m^2)`
= `(1)/(2sqrt(a^2 + m^2)) xx (0 + 2m) = m/sqrt(a^2 + m^2)`
and
`"dy"/"dm" = "d"/"dm"[log(a^2 + m^2)]`
= `(1)/(a^2 + m^2)."d"/"dm"(a^2 + m^2)`
= `(1)/(a^2 + m^2) xx (0 + 2m) = (2m)/(a^2 + m^2)`
∴ `"dy"/"dx" = (("dy"/"dm"))/(("dx"/"dm"))`
= `(((2m)/(a^2 + m^2)))/((m/sqrt(a^2 + m^2))`
= `(2)/sqrt(a^2 + m^2)`.
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