← Back to learning objective

Derivative at a point via limit definition

Let f(x)=x+4f(x) = \sqrt{x+4}. Use the limit definition of the derivative, f′(a)=lim⁡h→0f(a+h)−f(a)h,f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, to find f′(4)f'(4).

Hint

Write out f(4+h)=h+8f(4+h) = \sqrt{h+8} and f(4)=8f(4) = \sqrt{8}, form the difference quotient, then multiply numerator and denominator by the conjugate h+8+8\sqrt{h+8} + \sqrt{8} to eliminate the square roots before taking the limit.

Step-by-step solution

Start with the difference quotient: f(4+h)−f(4)h=h+8−8h.\frac{f(4+h)-f(4)}{h} = \frac{\sqrt{h+8} - \sqrt{8}}{h}. To simplify, multiply numerator and denominator by the conjugate h+8+8\sqrt{h+8}+\sqrt{8}, which turns the numerator into (h+8)−8=h(h+8) - 8 = h using the difference of squares.

This gives hh(h+8+8)=1h+8+8.\frac{h}{h(\sqrt{h+8}+\sqrt{8})} = \frac{1}{\sqrt{h+8}+\sqrt{8}}. Taking the limit as h→0h \to 0 yields f′(4)=18+8=128=142=28.f'(4) = \frac{1}{\sqrt{8}+\sqrt{8}} = \frac{1}{2\sqrt{8}} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap