I have thw function $$f(x) = (e^x+x^4)\cos(-2x) $$ which is defined for all $x \in \mathbb{R}$.

How do I prove it's differentiable for all x using the limit definition? I'm neither getting conclusive results with $$\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$ nor with $$\lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} $$ when I plug in the function for $f(x)$ and try to rearrange anf simplify it.

What am I missing?

Any help would be appreciated, thanks!

Edit: Is the trick here to Show that all 4 'partial' functions are differentiable on $\mathbb{R}$, thus making the composition of them differentiable in $\mathbb{R}$ as well?

  • 4
    $\begingroup$ @MikkoPesonen How do you get Taylor series if you don't even know whether the function is differentiable or not. $\endgroup$ – Fan May 16 '17 at 15:22
  • 1
    $\begingroup$ @MikkoPesonen This problem was asked before the introduction of taylor series, so it must be solveable without it. $\endgroup$ – Skydiver May 16 '17 at 15:23
  • $\begingroup$ The rules of differentiation exist for very good reasons. Why do you want to use the limit to prove the function is differentiable? $\endgroup$ – Umberto P. May 16 '17 at 15:25
  • $\begingroup$ You should have a review of the derivation of $(fg)'=f'g+fg'$ $\endgroup$ – Fan May 16 '17 at 15:25
  • $\begingroup$ @Fan I know how to differentiate it, but I have to prove that it's differentiable at all. $\endgroup$ – Skydiver May 16 '17 at 15:27

HINT: we have $$\frac{f(x)-f(x_o)}{x-x_o}=\frac{(e^x+x^4)\cos(-2x)-(e^{x_0}+x_0^4)\cos(-2x_0)}{x-x_0}$$ and use $$\cos(-x)=\cos(x)$$


I think this takes multiple applications of the product trick: If you need to work with $ab-cd$, it sometimes helps to transition from one product to the other one factor at a time:

$$ab - cd = ab -cb + cb -cd = (a-c)b + c(b-d).$$

You can multiply your function out and treat each sum separately. I'll do the first:

$$\frac{e^{x+h}\cos(-2(x+h)) - e^x\cos(-2x)}{h} $$

$$= \frac{e^{x+h}\cos(-2(x+h)) -e^{x}\cos(-2(x+h)) + e^{x}\cos(-2(x+h)-e^x\cos(-2x) }{h}$$

$$= \frac{\cos(-2(x+h))(e^{x+h}-e^x)}{h} + \frac{e^x(\cos(-2(x+h))-\cos(-2x))}{h}$$

You get two more, similar terms for the $x^4$ part. When you take the limit, everything works out swell.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.