Find dydx, if x = 1+u2, y = log(1 +u2) - Mathematics and Statistics

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Sum

Find `("d"y)/("d"x)`, if x = `sqrt(1 + "u"^2)`, y = log(1 +u2)

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Solution

x = `sqrt(1 + "u"^2)`

Differentiating both sides w.r.t. u, we get

`("d"x)/"du" = "d"/"du"(sqrt(1 + "u"^2))`

= `1/(2sqrt(1 + "u"^2))*"d"/"du"(1 + "u"^2)`

= `1/(2sqrt(1 + "u"^2)) xx (0 + 2"u")`

= `"u"/sqrt(1 + "u"^2)`

y = log(1 + u2)

Differentiating both sides w.r.t. u, we get

`("d"y)/"du" = "d"/"du"[log(1 + "u"^2)]`

= `1/(1 + "u"^2)*"d"/"du"(1 + "u"^2)`

= `1/(1 + "u"^2) xx (0 + 2"u") = (2"u")/(1 + "u"^2)`

∴ `("d"y)/("d"x) = ((("d"y)/"du"))/((("d"x)/("du"))) = (((2"u")/(1 + "u"^2)))/(("u"/sqrt(1 + "u"^2))`

= `2/(1 + "u"^2) xx sqrt(1 + "u"^2)`

∴ `("d"y)/("d"x) = 2/sqrt(1 + "u"^2)`

Concept: The Concept of Derivative - Derivatives of Logarithmic Functions
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Chapter 1.3: Differentiation - Q.4

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