Question

Differentiate this...
ASAP !​

Answer 1

Topic:

• Deferentiation.

Explanation:

Given:

$\bold{f(x) = {(1 + {x}^{2}) }^{ \frac{1}{2} }.} \\ \\ \bold{ suppose: \: (1 + {x}^{2}) = y. } \\ \\ \bold{differenciate \:f(y) \: with \: respect \: to \: x}. \\ \\ \bold{ \frac{df(y)}{dx} = 0 \times 1 {x}^{0 - 1} + 2 \times {x}^{2 - 1} .} \\ \\ \bold{ \implies{ \frac{df(y)}{dx} =2x \: - - - - - (1).}} \\ \\ \bold{and \: differenciate \:f(y) = {(y)}^{ \frac{1}{2} } \: with \: respect \: to \: y}. \\ \\ \bold{ \frac{df(y)}{dy} = \frac{1}{2} {(y)}^{ \frac{1}{2} - 1 }.} \\ \\ \bold{ \implies{\frac{df(y)}{dy} = \frac{1}{2} {(y)}^{ - \frac{1}{2} } = \frac{1}{2 \sqrt{y} } } \: - - - - - (2).} \\ \\ \bold{divide \: (1) \: by\: (2).then} \\ \\ \bold{ \implies{ \frac{df(y)}{dx} \times \frac{dy}{df(y)} = 2x \times \frac{1}{2 \sqrt{y} }. }} \\ \\ \bold{ \implies{ \frac{dy}{dx} = \frac{x}{ \sqrt{y} } }} \\ \bold{after \: putting \: all \: values \: in \: term \: of \: x \: it \: will \: become:} \\ \\ \bold{ \frac{df(x)}{d(x)} = \frac{x}{ \sqrt{1 + {x}^{2} } }. }$