I need some help with the derivative of this function.

Question

Hey guys i was wondering , what is the derivative function of this function.

f(x) = $\sqrt{x} - e^{-x}$

Any advise will be greated.

Answer 1

In general, $$\frac{d}{dx}(x^n) = nx^{n-1}$$ and $$\frac{d}{dx}(e^{ax}) = ae^{ax}.$$ Observing that $$f(x) = x^{\frac{1}{2}} - e^{-x}$$ we apply these general results to find that $$f'(x) = \frac {1}{2}x^{-\frac{1}{2}}+e^{-x}$$

Answer 2

$f(x)=\sqrt{x} -e^{-x}$
Let $p(x)=g(x)+h(x)$ it is true, that $p'(x)=g'(x)+h'(x)$ so we can write, that: \begin{eqnarray*} f(x)'&=&(\sqrt{x})'+(-e^{-x})'\\ (\sqrt{x})'&=&(x^{\frac{1}{2}})'=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\\ (-e^{-x})'&=&-1*(e^{-x})'=-1*e^{-x}*(-x)'=-1*e^{-x}*-1=e^{-x} \end{eqnarray*} Now sum it up and we get, that:
$$f'(x)=\frac{1}{2\sqrt{x}}+e^{-x}$$