## Calculus: Early Transcendentals (2nd Edition)

$a.$ The value of the derivative is $f'(1/4)=-4$ $b.$ The equation is $y=-4x+3$.
$a.$ By definition of a derivative at point $a=1/4$ we have $$f'(1/4)=\lim_{h\to0}\frac{f(1/4+h)-f(1/4)}{h}=\lim_{h\to0}\frac{\frac{1}{\sqrt{\frac{1}{4}+h}}-\frac{1}{\sqrt{\frac{1}{4}}}}{h}=\lim_{h\to0}\frac{\frac{1}{\sqrt{\frac{1+4h}{4}}}-\frac{1}{\frac{1}{2}}}{h}=\lim_{h\to0}\frac{\frac{1}{\frac{\sqrt{1+4h}}{2}}-2}{h}=\lim_{h\to0}\frac{\frac{2}{\sqrt{1+4h}}-2}{h}=\lim_{h\to0}\frac{\frac{2-2\sqrt{1+4h}}{\sqrt{1+4h}}}{h}=\lim_{h\to0}2\frac{1-\sqrt{1+4h}}{h\sqrt{1+4h}}=2\lim_{h\to0}\frac{1-\sqrt{1+4h}}{h\sqrt{1+4h}}\cdot\frac{1+\sqrt{1+4h}}{1+\sqrt{1+4h}}=2\lim_{h\to0}\frac{1^2-\sqrt{1+4h}^2}{h\sqrt{1+4h}(1+\sqrt{1+4h})}=2\lim_{h\to0}\frac{1-1-4h}{h(\sqrt{1+4h}+1+4h)}=2\lim_{h\to 0}\frac{-4h}{h(\sqrt{1+4h}+1+4h)}=2\lim_{h\to0}\frac{-4}{\sqrt{1+4h}+1+4h}=2\frac{-4}{\sqrt{1+4\cdot0}+1+4\cdot0}=-4.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=1/4$, $f(a)=1/\sqrt{1/4}=1/(1/2)=2$ and $f'(a)=-4$ we get $$y-2=-4(x-1/4)\Rightarrow y-2=-4x+1$$ which gives $$y=-4x+3.$$