Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To find:
The derivative of function f(x) using the first principle.
[tex]f(x)=\sqrt{x}[/tex]Solution:
By the first principle, the derivative of the function f(x) is given by:
[tex]f^{\prime}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}[/tex]So, the derivative of the given function can be obtained as follows:
[tex]\begin{gathered} f^{\prime}(x)=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \\ =\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\times\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ =\lim_{h\to0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} \\ =\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})} \\ =\lim_{h\to0}\frac{1}{(\sqrt{x+h}+\sqrt{x})} \\ =\frac{1}{\sqrt{x+0}+\sqrt{x}} \\ =\frac{1}{2\sqrt{x}} \end{gathered}[/tex]Thus, the derivative of the given function is:
[tex]f^{\prime}(x)=\frac{1}{2\sqrt{x}}[/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
