Show-that-g-f-x-g-f-x-f-x-

Question Number 94861 by Ar Brandon last updated on 21/May/20

Show that

{(g \circ f)}^{'} (x) = g^{'} (f (x)) \cdot f^{'} (x)

Commented by john santu last updated on 21/May/20

let f (x) = u \Rightarrow y = g (u)

chain rule

\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = g^{'} (u) \times f^{'} (x)

= g^{'} (f (x)) \times f^{'} (x)