To prove a function f(x) is differentiable in the domain of reals? Watch

monk1324
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Can you just differentiate it using normal rules (product, sum, chain etc) and show the derivative you get exists for all values of x out of reals?
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poorform
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Well if f,g are real valued functions and f and g are differentiable on R then the theorems of product rule, chain rule etc. say that the product of f and g or composition of f and g etc are also differentiable on R. So I suppose the way to go about it would be something like this example.

Claim: f(x)=sin(2x) is differentiable on R. Let f(x)=g(h(x)) where g(x)=sin(x) and h(x)=2x. Indeed g(x) is differentiable on R and moreover g'(x)=cos(x) also h(x)=2x is differentiable on R and moreover h'(x)=2 for all x in R.

So by the chain rule f(x)=g(h(x))=g o h is differentiable on R and moreover f'(x)=2cos(2x).

I hope that makes sense.

Obviously there may be a case where f is differentiable on all of R except some points in which case you may need to bring out the definition to see what happens at those points.
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Lord of the Flies
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(Original post by monk1324)
Can you just differentiate it using normal rules (product, sum, chain etc) and show the derivative you get exists for all values of x out of reals?
In general no - most differentiable functions aren't "expressible" in terms that allow you to differentiate using standard rules.

But even if a function is, the statement remains false - a standard example is f(x)=x^2\sin (x^{-1}). Differentiable everywhere, yet the derivative will appear to suggest otherwise.
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