use the definition of the derivative, which is $f(x)=lim_{h \to 0}\frac{f(x + h)-f(x)}{h}$ to find the…
use the definition of the derivative, which is $f(x)=lim_{h \to 0}\frac{f(x + h)-f(x)}{h}$ to find the derivative of each. show all work!! 1. $f(x)=x^{2}-4x$ 2. $f(x)=x^{3}+3x^{2}-4$ 3. $f(x)=\frac{1}{x}$ 4. $f(x)=sqrt{x}$ 5. $f(x)=\frac{x + 1}{x - 2}$
Answer
Explanation:
Step1: Find (f(x + h))
[ \begin{align*} f(x+h)&=(x + h)^2-4(x + h)\ &=x^{2}+2xh+h^{2}-4x-4h \end{align*} ]
Step2: Substitute into derivative formula
[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}\frac{(x^{2}+2xh+h^{2}-4x-4h)-(x^{2}-4x)}{h}\ &=\lim_{h\rightarrow0}\frac{x^{2}+2xh+h^{2}-4x-4h - x^{2}+4x}{h}\ &=\lim_{h\rightarrow0}\frac{2xh+h^{2}-4h}{h}\ &=\lim_{h\rightarrow0}(2x + h-4) \end{align*} ]
Step3: Evaluate the limit
[ \begin{align*} f^{\prime}(x)&=2x-4 \end{align*} ]
Answer: (f^{\prime}(x)=2x - 4)
Explanation:
Step1: Find (f(x + h))
[ \begin{align*} f(x+h)&=(x + h)^3+3(x + h)^2-4\ &=(x^{3}+3x^{2}h + 3xh^{2}+h^{3})+3(x^{2}+2xh+h^{2})-4\ &=x^{3}+3x^{2}h + 3xh^{2}+h^{3}+3x^{2}+6xh+3h^{2}-4 \end{align*} ]
Step2: Substitute into derivative formula
[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}\frac{(x^{3}+3x^{2}h + 3xh^{2}+h^{3}+3x^{2}+6xh+3h^{2}-4)-(x^{3}+3x^{2}-4)}{h}\ &=\lim_{h\rightarrow0}\frac{x^{3}+3x^{2}h + 3xh^{2}+h^{3}+3x^{2}+6xh+3h^{2}-4 - x^{3}-3x^{2}+4}{h}\ &=\lim_{h\rightarrow0}\frac{3x^{2}h + 3xh^{2}+h^{3}+6xh+3h^{2}}{h}\ &=\lim_{h\rightarrow0}(3x^{2}+3xh + h^{2}+6x+3h) \end{align*} ]
Step3: Evaluate the limit
[ \begin{align*} f^{\prime}(x)&=3x^{2}+6x \end{align*} ]
Answer: (f^{\prime}(x)=3x^{2}+6x)
Explanation:
Step1: Find (f(x + h))
[ f(x + h)=\frac{1}{x + h} ]
Step2: Substitute into derivative formula
[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{1}{x + h}-\frac{1}{x}}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{x-(x + h)}{x(x + h)}}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{x - x - h}{x(x + h)}}{h}\ &=\lim_{h\rightarrow0}\frac{-h}{xh(x + h)}\ &=\lim_{h\rightarrow0}\frac{-1}{x(x + h)} \end{align*} ]
Step3: Evaluate the limit
[ \begin{align*} f^{\prime}(x)&=-\frac{1}{x^{2}} \end{align*} ]
Answer: (f^{\prime}(x)=-\frac{1}{x^{2}})
Explanation:
Step1: Find (f(x + h))
[ f(x + h)=\sqrt{x + h} ]
Step2: Substitute into derivative formula
[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}\frac{\sqrt{x + h}-\sqrt{x}}{h}\ &=\lim_{h\rightarrow0}\frac{\sqrt{x + h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x + h}+\sqrt{x}}{\sqrt{x + h}+\sqrt{x}}\ &=\lim_{h\rightarrow0}\frac{(x + h)-x}{h(\sqrt{x + h}+\sqrt{x})}\ &=\lim_{h\rightarrow0}\frac{h}{h(\sqrt{x + h}+\sqrt{x})}\ &=\lim_{h\rightarrow0}\frac{1}{\sqrt{x + h}+\sqrt{x}} \end{align*} ]
Step3: Evaluate the limit
[ \begin{align*} f^{\prime}(x)&=\frac{1}{2\sqrt{x}} \end{align*} ]
Answer: (f^{\prime}(x)=\frac{1}{2\sqrt{x}})
Explanation:
Step1: Find (f(x + h))
[ \begin{align*} f(x + h)&=\frac{(x + h)+1}{(x + h)-2}=\frac{x+h + 1}{x+h - 2} \end{align*} ]
Step2: Substitute into derivative formula
[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{x+h + 1}{x+h - 2}-\frac{x + 1}{x - 2}}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{(x+h + 1)(x - 2)-(x + 1)(x+h - 2)}{(x+h - 2)(x - 2)}}{h}\ &=\lim_{h\rightarrow0}\frac{(x+h + 1)(x - 2)-(x + 1)(x+h - 2)}{h(x+h - 2)(x - 2)}\ &=\lim_{h\rightarrow0}\frac{x^{2}-2x+hx-2h+x - 2-(x^{2}+xh-2x+x+h - 2)}{h(x+h - 2)(x - 2)}\ &=\lim_{h\rightarrow0}\frac{x^{2}-2x+hx-2h+x - 2 - x^{2}-xh + 2x - x - h+2}{h(x+h - 2)(x - 2)}\ &=\lim_{h\rightarrow0}\frac{-3h}{h(x+h - 2)(x - 2)}\ &=\lim_{h\rightarrow0}\frac{-3}{(x+h - 2)(x - 2)} \end{align*} ]
Step3: Evaluate the limit
[ \begin{align*} f^{\prime}(x)&=-\frac{3}{(x - 2)^{2}} \end{align*} ]