Suppose $f$ and $g$ are non-constant, differentiable, real- valued functions defined on $(-\infty, \infty) .$ Furthermore, suppose that for each pair of real numbers $x$ and $y$

$f(x+y)=f(x) f(y)-g(x) g(y)$ and

$g(x+y)=f(x) g(y)+g(x) f(y)$

If $$f^{\prime}(0)=0,$ prove that $(f(x))^{2}+(g(x))^{2}=1$ for all $x$$

Proof inside

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Numerade Educator

Campbell University

University of Nottingham

Boston College

{'transcript': "okay. And this problem capital F of X is f of X to the Alfa and we want to find f prime. So we're going to use the chain rule. The outside function is the F function. Lower case F. So it's derivative would be lower case f prime of X of out extra. The Alfa Times, the derivative of the inside the anti function is extra the Alfa and we can use the power rule on that Bring down the Alfa and then raise X to the Alfa minus one for G of X capital G of X is f of X to the Alfa Power. So we have similar functions but in a different order and we want g prime So again using the chain rule. Now the outside function is the Alfa power function. So we bring down the Alfa and we raise F of X to the Alfa minus one and now for the derivative of the inside, the derivative of F of X would be f prime of X"}

Oregon State University