How do you find f'(x) using the definition of a derivative for f(x)=x+x+7?

Explanation:f(x)=x+x+7dydx=1+12xAnswer:dydx=1+12x

f′(x)=limh→0f(x+h)−f(x)hlimh→0x+h+x+h+7−x−x−7h=limh→0h+x+h−xh=limh→0(1+x+h−xh)==1+limh→0x+h−xhx+h+xx+h+x==1+limh→0x+h−xh(x+h+x)==1+limh→0hh(x+h+x)=1+limh→01x+h+x=1+1x+x=1+12x

Explanation:f(x)=x+x+7dydx=1+12xAnswer:dydx=1+12x

f′(x)=limh→0f(x+h)−f(x)hlimh→0x+h+x+h+7−x−x−7h=limh→0h+x+h−xh=limh→0(1+x+h−xh)==1+limh→0x+h−xhx+h+xx+h+x==1+limh→0x+h−xh(x+h+x)==1+limh→0hh(x+h+x)=1+limh→01x+h+x=1+1x+x=1+12x

"How do you find f'(x) using the definition..." was answered by our experts

Emilia Hoffman

Answered question

2022-04-15

How do you find f'(x) using the definition of a derivative for

f (x) = x + \sqrt{x} + 7

Answer & Explanation

analiticozuod

Beginner2022-04-16Added 10 answers

Explanation:

f (x) = x + \sqrt{x} + 7

\frac{d y}{d x} = 1 + \frac{1}{2 \sqrt{x}}

Answer:

\frac{d y}{d x} = 1 + \frac{1}{2 \sqrt{x}}

Naima Silva

Beginner2022-04-17Added 9 answers

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

lim_{h \to 0} \frac{x + h + \sqrt{x + h} + 7 - x - \sqrt{x} - 7}{h} =

lim_{h \to 0} \frac{h + \sqrt{x + h} - \sqrt{x}}{h} = lim_{h \to 0} (1 + \frac{\sqrt{x + h} - \sqrt{x}}{h}) =

= 1 + lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} =

= 1 + lim_{h \to 0} \frac{x + h - x}{h (\sqrt{x + h} + \sqrt{x})} =

= 1 + lim_{h \to 0} \frac{h}{h (\sqrt{x + h} + \sqrt{x})} = 1 + lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}

= 1 + \frac{1}{\sqrt{x} + \sqrt{x}} = 1 + \frac{1}{2 \sqrt{x}}

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