Derivatives question help The question is :Find the derivative of $f(x)=e^c + c^x$. Assume that c is a constant.
Wouldn't $f'(x)=   ce^{c-1} + xc^{x-1}$.    It keeps saying this answer is incorrect, What am i doing wrong?
 A: Consider $$f(x)=e^{c}+c^{x}$$ where $c$ is a constant. We know that since $c$ is a constant, $e^c$ is also a constant making ${d\over dx}(e^c)=0$. Also, ${d\over dx}(c^x)=c^x\ln c$. The reason for this is because ${d\over dx}(c^x)={d\over dx}{(e^{\ln c})^x}={d\over dx}({e^{{(\ln c}){x}})}=e^{({\ln c)} x}\cdot {d\over dx} {(\ln c)}x=(e^{\ln c})^x\cdot (\ln c)=c^x\ln c$. Thus $$f'(x)={c^x\ln c}.$$ 
A: There are distinct derivation rules for two cases:


*

*A variable raised to a constant

*A constant raised to a variable.


You tried to apply the first case to the second term, $c^x$, where you really want the second case. Consider the general rule for constants raised to variables, $$ \frac{d}{dx} (a^u) = a^u \ln a \frac{du}{dx}. $$
Regarding the first term $e^c$, perhaps you missed that $e$ is a mathematical constant: the base of the natural logarithm, approximately equal to 2.71828.
A: Since "c" is a constant,the derivative of the first term is zero and then the derivative of f(x) is the same as the derivative of g(x)=c^x. I suggest you go through loagrithms and re-express g(x) in a more familiar form. Are you able to continue with this ?
A: As stated in the comments, $e^c$ is a constant, so its derivative is zero. As for the second term, $c^x$, the rule for the derivative of a variables as an exponents is better understood if you write the function as
$c^x = e^{x \ln{c}}$
So, clearly the derivative is equal to $c^x\ln{c} $
To sum up, $f'(x) = c^x \ln{c}$
