Question

Find `"dy"/"dx"`, if : x = `sqrt(a^2 + m^2), y = log(a^2 + m^2)`

Answer 1

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)`.