If $f(x+y)=f(x)\cdot f(y) \text{ } \forall\text{ } x, y\text{ } \in R \text{ and } f(5) = 2, f'(0) = 3; \text{ find } f'(5)$

My question:

We know that such types of functions are the exponential functions $f(x) = a^x$. So, $f'(x) = x\cdot a^{x-1}$. So, $f'(0)$ should be $0\cdot a^{-1}$, but it is given $3$.

I don't want the solution to the question given. I want the explanation behind why my working seems to be producing weird results.

UPDATE: Since I posted this question, I realized there's two methods to solve this question. One clearly shown by Emilio below, the other that my textbook does (which does NOT use the fact that f(x) is exponential fn.). Now, interestingly enough, the answer posted below clearly marks the question wrong, while my textbook elegantly solves it.

So, my new question is:

Why those two methods diverge so much, why don't we get the same result for the same question by applying either of them?

My textbook's method for reference:

Let $x=0, y=5$. Putting in given equation implies $f(0) =1$.

Now, $$f'(5)=\lim_{h\to 0}\frac{f(5+h)-f(5)}{h}=\lim_{h\to 0}\frac{f(5)f(h)-f(5)}{h}=f(5)\lim_{h\to 0}\frac{f(h)-1}{h}=f(5)\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=f(5)\cdot f'(0)=2*3=6$$

Final update: So, as stated in comments by dxiv, my textbook solves the problem, but doesn't prove if the function exists in the first place. That's the first time I saw such a thing. I've accepted the current answer because it shows why the question is wrong. Thanks everyone!

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    $\begingroup$ $f'(x)=a^x\ln a$ $\endgroup$ – kingW3 Dec 18 '16 at 12:24
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    $\begingroup$ $f'(x)=a^x\ln a\neq xa^{x-1}$. $\endgroup$ – Rohan Dec 18 '16 at 12:25
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    $\begingroup$ @GaurangTandon Rewrite $a^x = e^{x\ln a} = g(x\ln a)$ for $g=\exp$. Differentiating with respect to $x$, with the chain rule, you get $f'(x) = \frac{d}{dx}(x\ln a)\cdot g'(x\ln a) = \ln a \cdot e^{x\ln a} = (\ln a) a^x$. $\endgroup$ – Clement C. Dec 18 '16 at 12:30
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    $\begingroup$ Just a comment, but I assume you're aware you don't need to use the result that such an $f$ is an exponential function in order to answer this question. $\endgroup$ – Dan Rust Dec 18 '16 at 12:55
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    $\begingroup$ Are you sure of the problem? From $f'(0)=3$ we find $f(x)=e^{3x}$ so $f(5)\ne 2$. $\endgroup$ – Emilio Novati Dec 18 '16 at 13:08

As a first step, as noted in the comments, you have a mistake in the derivative. The function $y=a^x$, can be written as $y=(e^{\ln a})^x=e^{x \ln a} $ , so the derivative ( using the chain rule) is $y'=e^{x \ln a}\cdot \ln a =a^x\ln a$.

The condition $y'(0)=3$ gives $\ln a= 3$ so the function is $y=e^{3x}$ and , for $x=5$, this gives $y=e^{15}\ne 2$.

  • $\begingroup$ Does that mean the question is wrong? $\endgroup$ – Gaurang Tandon Dec 21 '16 at 10:55
  • $\begingroup$ Yes, there is something wrong in the question. $\endgroup$ – Emilio Novati Dec 21 '16 at 13:32

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