3) Why? Proof: Let y = f1(x) + f2(x), Where f1(x) and f2(x) are differentiable functions of x. Then Thus . 4) Why? Proof: Let y = f1(x).f2(x) where both f1(x) and f2(x) are differentiable function of x. 5) Why? Proof: Let , where f1(x) and f2(x) are differentiable functions at all points in its domain and f2(x) ¹ 0. 6) Why? Proof: Let y = f (t), t = g(x) Then y = f (g(x)) is a function of x. Dy = f (t+Dt) - f (t) And Dt = g (x+Dx) - g(x) Assuming that for sufficient small values of Dx, Dt ¹ 0 this will necessarily be the case if then since g is differentiable it is continuous and hence when Dx®0, x+Dx®x any g(x+Dx) ® g(x). Therefore t+Dt®t and Dt®0, since Dt ¹ 0 and Dx®0 we may write, . Clearly f and g are both continuous functions because they are differentiable thus Dt®0 when Dx®0 and Dy®0 when Dt®0. Hence Standard Formulae of Differentiation: 1) Proof: Let Thus 2) Proof: Thus 3) Proof: So, 4) Proof:
|