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Question: How do you find \(f'(x)\) using the definition of a derivative for \(f(x)=\sqrt{1+2x}?\)...

How do you find f′(x)f'(x) using the definition of a derivative for f(x)=1+2x?f(x)=\sqrt{1+2x}?

Explanation

Solution

We know that the term derivative characterises the rate of change of function. Bu calculating the ratio of change of function Δy\Delta y to the ratio of change of independent variable Δx\Delta x So, here in the definition of derivative ratio is consider in limit as Δx→0.\Delta x\to 0. It is represent in the formula as:
f′(x0)=lim⁡Δx→0 ΔyΔxf'\left( {{x}_{0}} \right)=\underset{\Delta x\to 0}{\mathop{\lim }}\,\dfrac{\Delta y}{\Delta x}
=lim⁡Δx→0 f(x0+Δx)−f(x0)Δx=\underset{\Delta x\to 0}{\mathop{\lim }}\,\dfrac{f\left( {{x}_{0}}+\Delta x \right)-f\left( {{x}_{0}} \right)}{\Delta x}
Where as f(x)f(x) be a function is said to differential at x0{{x}_{0}}
As in the given question we have used the definition of derivative. After that we have to simplify the quotient and contently Δx\Delta x if this is possible, then find the derivative of f′(x)f'(x) then apply the limit to the quotient. If limit exist then function f(x)f(x) is differentiable at x0.{{x}_{0}}.

Complete step-by-step answer:
Let,
f(x)=1+2xf(x)=\sqrt{1+2x}
Then the derivative at x=ax=a is defined as a limit.
f′(a)=lim⁡h→0 f(a+h)−f(a)hf'(a)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(a+h)-f\left( a \right)}{h}
=lim⁡h→0 1+2(a+h)−1+2ah=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+2\left( a+h \right)}-\sqrt{1+2a}}{h}
Multiplying out the numerator we get,
=lim⁡h→0 1+2(a+b)−1+2ah×1+2(a+h)+1+2a1+2(a+b)+1+2a=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+2\left( a+b \right)}-\sqrt{1+2a}}{h}\times \dfrac{\sqrt{1+2\left( a+h \right)}+\sqrt{1+2a}}{\sqrt{1+2\left( a+b \right)}+\sqrt{1+2a}}
=lim⁡h→0 1+2(a+h)−1+2ah×1+2(a+b)+1+2a1+2(a+h)+2a=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+2\left( a+h \right)}-\sqrt{1+2a}}{h}\times \dfrac{\sqrt{1+2\left( a+b \right)}+\sqrt{1+2a}}{\sqrt{1+2\left( a+h \right)}+\sqrt{2a}}
Cancelling the under root, we get,
=lim⁡h→0 [1+2(a+h)]−[a+2a]h(1+2(a+h))+(1+2a)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left[ 1+2\left( a+h \right) \right]-\left[ a+2a \right]}{h\left( \sqrt{1+2\left( a+h \right)} \right)+\left( \sqrt{1+2a} \right)}
Then,
Cancelling, ′h′'h' from numerator and denominator we get,
=lim⁡h→0 2hh(1+2(a+h)+1+2a)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2h}{h\left( \sqrt{1+2\left( a+h \right)}+\sqrt{1+2a} \right)}
=lim⁡h→0 21+2(a+h)+1+2a=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2}{\sqrt{1+2\left( a+h \right)}+\sqrt{1+2a}}
=21+2a+1+2a=\dfrac{2}{\sqrt{1+2a}+\sqrt{1+2a}}
f′(x)=11+2af'(x)=\dfrac{1}{\sqrt{1+2a}}

By using the definition of a derivative for f(x)=1+2xf(x)=\sqrt{1+2x} we get, f′(x)=11+2af'(x)=\dfrac{1}{\sqrt{1+2a}}.

Additional Information:
The term derivative is the rate of change of a function with respect to a variable. It is fundamental to the solution of problems in differential equations. As for understanding we have to take as a example the derivative of f(x)f(x) at x0,{{x}_{0}}, can be written as f′(x0)[dfdx](x0)f'({{x}_{0}})\left[ \dfrac{df}{dx} \right]\left( {{x}_{0}} \right).
Another term in differentiation, is the rate of change of a function or it is a process of finding derivatives. There is another basic rule, which we also called it as a chain rule. In that you have to provide a way to differentiate a composite function [it can be written inside another function].

Note:
Before solving any problem first we have to check the type of problem and what it is asking. Sometimes we directly solve it which is incorrect and also these are various rules applied in various problems. After solving, analyse the problem whether the written derivative is correct or not.