हिंदी

If x = 1+u2, y = log(1+u2), then find dydx.

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प्रश्न

If x = `sqrt(1 + u^2)`, y = `log(1 + u^2)`, then find `(dy)/(dx).`

योग
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उत्तर

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

Differentiating w.r.t.. 'u',

`(dx)/(du) = (2u)/(2sqrt(1 + u^2)`

= `u/sqrt(1 + u^2)`  ......(1)

Now, y = `log(1 + u^2)`

Differentiating w.r.t, u,

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

⇒ `(dy)/(du) = (2u)/(1 + u^2)`   ......(2)

We have,

`(dy)/(dx) = ((dy)/(du))/((dx)/(du))`

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

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

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

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