Using the given definition, for [tex]f(x)=\frac1{\sqrt x}+x[/tex], we have
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\left(\frac1{\sqrt{x+h}}+x+h\right)-\left(\frac1{\sqrt x}+x\right)}h[/tex]
Right away, we see x and -x in the numerator, so we can drop those terms.
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}+h-\frac1{\sqrt x}}h[/tex]
Remember that limits distribute over sums, i.e.
[tex]\displaystyle\lim_{x\to c}(f(x)+g(x))=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)[/tex]
so we can separate the h from everything else in the numerator:
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}-\frac1{\sqrt x}}h+\lim_{h\to0}\frac hh[/tex]
Since h ≠ 0, we have [tex]\frac hh=1[/tex], so the second limit is simply 1.
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}-\frac1{\sqrt x}}h+1[/tex]
For the remaining limit, focus on the numerator for now. Combine the fractions in the numerator:
[tex]\dfrac1{\sqrt{x+h}}-\dfrac1{\sqrt x}=\dfrac{\sqrt x-\sqrt{x+h}}{\sqrt x\sqrt{x+h}}[/tex]
Recall the difference of squares identity,
[tex]a^2-b^2=(a-b)(a+b)[/tex]
Let [tex]a=\sqrt x[/tex] and [tex]b=\sqrt{x+h}[/tex]. Multiply the numerator and denominator by [tex](a+b)[/tex], so that the numerator can be condensed using the identity above.
[tex]\dfrac{\sqrt x-\sqrt{x+h}}{\sqrt x\sqrt{x+h}}\cdot\dfrac{\sqrt x+\sqrt{x+h}}{\sqrt x+\sqrt{x+h}}[/tex]
[tex]=\dfrac{(\sqrt x)^2-(\sqrt{x+h})^2}{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]
[tex]=\dfrac{x-(x+h)}{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]
[tex]=-\dfrac h{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]
Back to the limit: all this rewriting tells us that
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{-\frac h{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}}h+1[/tex]
Again, the h's cancel, and we can pull out the factor of -1 from the numerator and simplify the fraction:
[tex]f'(x)=\displaystyle-\lim_{h\to0}\frac1{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}+1[/tex]
The remaining expression is continuous at h = 0, so we can evaluate the limit by substituting directly:
[tex]f'(x)=-\dfrac1{\sqrt x\sqrt{x+0}(\sqrt x+\sqrt{x+0})}+1[/tex]
[tex]f'(x)=-\dfrac1{2x\sqrt x}+1[/tex]
or, if we write [tex]\sqrt x=x^{1/2}[/tex], we get
[tex]f'(x)=-\dfrac12x^{-3/2}+1[/tex]
