The Definition of the Derivative — Question 10

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Question 10

Let f(x)=x+3f(x) = \sqrt{x + 3} Use the definition of the derivative to find f′(x)f'(x). Do not apply any derivative shortcuts — use first principles only.

Original worksheet page 1: question and worked solution for 3-1-010
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Question 10 - Solution

We are given: f(x)=x+3f(x) = \sqrt{x + 3}

Apply the definition of the derivative: f′(x)=limh→0f(x+h)−f(x)h=limh→0x+h+3−x+3hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h}

To evaluate this limit, multiply numerator and denominator by the conjugate of the numerator: =limh→0x+h+3−x+3h⋅x+h+3+x+3x+h+3+x+3= \lim_{h \to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h} \cdot \frac{\sqrt{x + h + 3} + \sqrt{x + 3}}{\sqrt{x + h + 3} + \sqrt{x + 3}}

Use the identity (a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2: =limh→0(x+h+3)−(x+3)h(x+h+3+x+3)=limh→0hh(x+h+3+x+3)= \lim_{h \to 0} \frac{(x + h + 3) - (x + 3)}{h(\sqrt{x + h + 3} + \sqrt{x + 3})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h + 3} + \sqrt{x + 3})}

Cancel hh in numerator and denominator: =limh→01x+h+3+x+3= \lim_{h \to 0} \frac{1}{\sqrt{x + h + 3} + \sqrt{x + 3}}

Now take the limit as h→0h \to 0: f′(x)=12x+3f'(x) = \frac{1}{2\sqrt{x + 3}}

Final Answer: f′(x)=12x+3f'(x) = \boxed{\frac{1}{2\sqrt{x + 3}}}

Original worksheet page 2: question and worked solution for 3-1-010

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