Question: User: Top-left: Let $f(x)=\sqrt{5x+8}$. According to the definition,
Top-left: $f'(x)=\lim_{h\to 0}$ [blank input box] (Enter the expression in reduced form)
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Bottom-left: Question 10
Bottom-left: Let $f(x)=\dfrac{1}{5x-4}$. According to the definition,
Bottom-left: $f'(x)=\lim_{h\to 0}$ [blank input box] (Enter the limit in reduced form)please solve
1. **State the problem:**
We are given the function $f(x)=\dfrac{1}{5x-4}$ and asked to find its derivative $f'(x)$ using the definition of the derivative.
2. **Recall the definition of the derivative:**
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
3. **Apply the definition to the given function:**
$$f'(x) = \lim_{h \to 0} \frac{\frac{1}{5(x+h)-4} - \frac{1}{5x-4}}{h}$$
4. **Simplify the numerator:**
$$= \lim_{h \to 0} \frac{\frac{1}{5x+5h-4} - \frac{1}{5x-4}}{h}$$
5. **Find common denominator for the numerator:**
$$= \lim_{h \to 0} \frac{\frac{5x-4 - (5x+5h-4)}{(5x+5h-4)(5x-4)}}{h}$$
6. **Simplify the numerator inside the fraction:**
$$= \lim_{h \to 0} \frac{\frac{5x-4 - 5x - 5h + 4}{(5x+5h-4)(5x-4)}}{h} = \lim_{h \to 0} \frac{\frac{-5h}{(5x+5h-4)(5x-4)}}{h}$$
7. **Rewrite the complex fraction:**
$$= \lim_{h \to 0} \frac{-5h}{h (5x+5h-4)(5x-4)}$$
8. **Cancel $h$ in numerator and denominator:**
$$= \lim_{h \to 0} \frac{-5\cancel{h}}{\cancel{h} (5x+5h-4)(5x-4)} = \lim_{h \to 0} \frac{-5}{(5x+5h-4)(5x-4)}$$
9. **Evaluate the limit as $h \to 0$:**
$$= \frac{-5}{(5x-4)^2}$$
**Final answer:**
$$f'(x) = \frac{-5}{(5x-4)^2}$$