Find dydx, if x = em, y = em Solution: Given, x = em and y = em Now, y = em Diff.w.r.to m, dydm=emd□dm ∴ dydm=em⋅12m .....(i) Now, x = em Diff.w.r.to m, dxdm=□ .....(ii) Now, dydx=dydx□ ∴ dydx=em□em - Mathematics and Statistics

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Sum

Find `("d"y)/("d"x)`, if x = em, y = `"e"^(sqrt("m"))`

Solution: Given, x = em and y = `"e"^(sqrt("m"))`

Now, y = `"e"^(sqrt("m"))`

Diff.w.r.to m,

`("d"y)/"dm" = "e"^(sqrt("m"))("d"square)/"dm"`

∴ `("d"y)/"dm" = "e"^(sqrt("m"))*1/(2sqrt("m"))`    .....(i)

Now, x = em

Diff.w.r.to m,

`("d"x)/"dm" = square`    .....(ii)

Now, `("d"y)/("d"x) = (("d"y)/("d"m))/square`

∴ `("d"y)/("d"x) = (("e"sqrt("m"))/square)/("e"^"m")`

∴  `("d"y)/("d"x) = ("e"^(sqrt("m")))/(2sqrt("m")*"e"^("m")`

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Solution

Given, x = em and y = `"e"^(sqrt("m"))`

Now, y = `"e"^(sqrt("m"))`

Diff.w.r.to m,

`("d"y)/"dm" = "e"^(sqrt("m"))("d"sqrt("m"))/"dm"`

∴ `("d"y)/"dm" = "e"^(sqrt("m"))*1/(2sqrt("m"))`    .....(i)

Now, x = em

Diff.w.r.to m,

`("d"x)/"dm"` = em    .....(ii)

Now, `("d"y)/("d"x)` = `(("d"y)/("d"m))/(("d"x)/("dm"))`

∴ `("d"y)/("d"x) = (("e"sqrt("m"))/(2sqrt("m")))/("e"^"m")`

∴  `("d"y)/("d"x) = ("e"^(sqrt("m")))/(2sqrt("m")*"e"^("m")`

Concept: Derivatives of Parametric Functions
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Chapter 1.3: Differentiation - Q.6

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