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"` &there4; `("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` &there4; `("d"y)/("d"x) = (("e"sqrt("m"))/square)/("e"^"m")` &there4; `("d"y)/("d"x) = ("e"^(sqrt("m")))/(2sqrt("m")*"e"^("m")`

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

Fill in the Blanks

Sum

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

Solution: Given, x = e^m 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 = e^m

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")`

Given, x = e^m 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 = e^m

Diff.w.r.to m,

`("d"x)/"dm"` = e^m .....(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

Is there an error in this question or solution?

Chapter 1.3 Differentiation
Q.6 | Q 4