A Arithela Member Joined Jun 23, 2007 Messages 306 Gender Female HSC 2008 May 5, 2008 #1 diferentiate sin[e^(2x)] help please
foram Awesome Member Joined Mar 18, 2008 Messages 1,015 Location Beyond Godlike Gender Male HSC 2009 May 5, 2008 #2 Arithela said: diferentiate sin[e^(2x)] help please Click to expand... 2e^2x .cos(e^2x) I think, using chain rule.
Arithela said: diferentiate sin[e^(2x)] help please Click to expand... 2e^2x .cos(e^2x) I think, using chain rule.
A Arithela Member Joined Jun 23, 2007 Messages 306 Gender Female HSC 2008 May 5, 2008 #3 thats right, show me working please
P pLuvia Guest May 5, 2008 #4 Let g(x) and f(y) be functions d/dx(g(f(y)))=f'(y)g'([f(y)]) Hence d/dx(sin[e^(2x)]) =2e2x*cos[e2x] That's the only working you need
Let g(x) and f(y) be functions d/dx(g(f(y)))=f'(y)g'([f(y)]) Hence d/dx(sin[e^(2x)]) =2e2x*cos[e2x] That's the only working you need
foram Awesome Member Joined Mar 18, 2008 Messages 1,015 Location Beyond Godlike Gender Male HSC 2009 May 5, 2008 #5 y=sin(e^2x) y'= dy/du . du/dx (chain rule) let u=e^2x dy/du = cos u du/dx = 2e^2x dy/du . du/dx = 2e^2x . cos(u) = 2e^2x cos(e^2x)
y=sin(e^2x) y'= dy/du . du/dx (chain rule) let u=e^2x dy/du = cos u du/dx = 2e^2x dy/du . du/dx = 2e^2x . cos(u) = 2e^2x cos(e^2x)
Mark576 Feel good ... Joined Jul 6, 2006 Messages 515 Gender Male HSC 2008 May 5, 2008 #6 d/dx[sin(f(x))] = f'(x).cos(f(x)) and if f(x) = e2x then f'(x) = 2e2x So then d/dx(sin(e2x)) = 2e2x.cos(e2x)
d/dx[sin(f(x))] = f'(x).cos(f(x)) and if f(x) = e2x then f'(x) = 2e2x So then d/dx(sin(e2x)) = 2e2x.cos(e2x)