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Non-integer power rule by definition

mark

Use the definition of the derivative to find the derivative of

f(x) = \frac{1}{\sqrt{x}}.
mearing

Here’s my best shot
\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}
\lim_{h \to 0}\frac{\frac{\sqrt{x}(1)}{\sqrt{x}\sqrt{x+h}}-\frac{\sqrt{x+h}(1)}{\sqrt{x}\sqrt{x+h}}}{h}
\lim_{h \to 0}\frac{\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}}{h}
\lim_{h \to 0}\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}*\frac{1}{h}
\lim_{h \to 0}\frac{1}{h}*\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}*\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}
\lim_{h \to 0}\frac{1}{h}*\frac{(\sqrt{x}-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}
\lim_{h \to 0}\frac{1}{h}*\frac{x-(x+h)}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}
\lim_{h \to 0}\frac{1}{h}*\frac{x-x-h}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}
\lim_{h \to 0}\frac{1}{h}*\frac{-h}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}
\lim_{h \to 0}\frac{-1}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}
\lim_{h \to 0}\frac{-1}{(\sqrt{x}\sqrt{x+0})(\sqrt{x}+\sqrt{x+0})}
f'(x)=\frac{-1}{(\sqrt{x(x})(\sqrt{x}+\sqrt{x})}
f'(x)=\frac{-1}{(\sqrt{x^2})(\sqrt{x}+\sqrt{x}))}
f'(x)=\frac{-1}{(x\sqrt{x}+x\sqrt{x})}
f'(x)=\frac{-1}{2x\sqrt{x}}
f'(x)=-\frac{1}{2}x^{-1}x^{-\frac{1}{2}}

f'(x)=-\frac{1}{2}x^{-\frac{3}{2}}

jbarry1

Sorry about the number of brackets, but here is my go at it:

f^{\prime }(x)=\frac{\left(f(x+h)-f(x)\right)}{h}
f^{\prime }(x)=\frac{\left(\left(\frac{1}{\sqrt{x+h}}\right)-\left(\frac{1}{\sqrt{x}}\right)\right)}{h}
f^{\prime }(x)=\frac{1}{h}\left(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right)
f^{\prime }(x)=\frac{1}{h}\left(\frac{\left(\sqrt{x}-\sqrt{x+h}\right)}{\sqrt{x}\left(\sqrt{x+h}\right)}\right)
f^{\prime }(x)=\frac{1}{h}\left(\frac{\left(\sqrt{x}-\sqrt{x+h}\right)}{\sqrt{x}\left(\sqrt{x+h}\right)}\right)\left(\frac{\left(\sqrt{x}+\sqrt{x+h}\right)}{\sqrt{x}+\sqrt{x+h}}\right)
f^{\prime }(x)=\frac{1}{h}\left(\frac{\left(x-\left(x+h\right)\right)}{\left(\sqrt{x}\left(\sqrt{x+h}\right)\right)\left(\sqrt{x}+\sqrt{x+h}\right)}\right)
f^{\prime }(x)=\frac{1}{h}\left(\frac{-h}{\left(\sqrt{x}\left(\sqrt{x+h}\right)\right)\left(\sqrt{x}+\sqrt{x+h}\right)}\right)
f^{\prime }(x)=\frac{h}{h}\left(\frac{-1}{\left(\sqrt{x}\left(\sqrt{x+h}\right)\right)\left(\sqrt{x}+\sqrt{x+h}\right)}\right)
f^{\prime }(x)=\frac{-1}{\left(\sqrt{x}\left(\sqrt{x+h}\right)\right)\left(\sqrt{x}+\sqrt{x+h}\right)}
f^{\prime }(x)=\lim _{h\to 0}\left(\frac{-1}{\left(\sqrt{x}\left(\sqrt{x+0}\right)\right)\left(\sqrt{x}+\sqrt{x+0}\right)}\right)=\frac{-1}{x\left(2\sqrt{x}\right)}=\frac{-1}{x^12x^{\left(\frac{1}{2}\right)}}=\frac{-1}{2x^{\left(\frac{3}{2}\right)}}

I thought that it would be okay to leave it in this fraction form to end? I realise the x could be moved to the top making its power negative but wondered if that was a big deal or not.