I don't even know how to start this problem. g'(x) = g'(0) = Please help....
M Marcia New member Joined Oct 18, 2005 Messages 14 Oct 18, 2005 #1 I don't even know how to start this problem. g'(x) = g'(0) = Please help....
T tridelta246 New member Joined Oct 17, 2005 Messages 14 Oct 18, 2005 #2 Re: Let f be differentiable at 0 let g(x) = f(x^2). Find g' Marcia said: I don't even know how to start this problem. g'(x) = g'(0) = Please help.... Click to expand... Use the chain rule when g(x) = f(h(x)): g'(x) = dg/dx = d{f(h(x))}/dx = (df/dh)*(dh/dx) where for this case: g(x) = f(x^2) = f(h(x)) h(x) = x^2 df/dh = f'(x) dh/dx = 2*x Thus: g'(x) = {f'(x)}*(2*x) ..
Re: Let f be differentiable at 0 let g(x) = f(x^2). Find g' Marcia said: I don't even know how to start this problem. g'(x) = g'(0) = Please help.... Click to expand... Use the chain rule when g(x) = f(h(x)): g'(x) = dg/dx = d{f(h(x))}/dx = (df/dh)*(dh/dx) where for this case: g(x) = f(x^2) = f(h(x)) h(x) = x^2 df/dh = f'(x) dh/dx = 2*x Thus: g'(x) = {f'(x)}*(2*x) ..
