How do you find f'(x) using the definition of a derivative for $f (x) = x + \sqrt{x} + 7$ $f(x)=x + sqrtx + 7$ ?

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Answer 1 · 2015-09-27T04:27:43

$\frac{d y}{d x} = 1 + \frac{1}{2 \sqrt{x}}$ $dy/dx=1+1/(2sqrtx)$

$f (x) = x + \sqrt{x} + 7$ $f(x)=x+sqrtx+7$
$\frac{d y}{d x} = 1 + \frac{1}{2 \sqrt{x}}$ $dy/dx=1+1/(2sqrtx)$

Answer 2 · 2015-09-27T04:53:35

See the explanation.

$f'(x)=lim_(h->0) (f(x+h)-f(x))/h$

$lim_(h->0) (x+h+sqrt(x+h)+7-x-sqrtx-7)/h=$

$lim_(h->0) (h+sqrt(x+h)-sqrtx)/h=lim_(h->0)(1+(sqrt(x+h)-sqrtx)/h)=$

$=1+lim_(h->0) (sqrt(x+h)-sqrtx)/h (sqrt(x+h)+sqrtx)/(sqrt(x+h)+sqrtx)=$

$=1+lim_(h->0) (x+h-x)/(h(sqrt(x+h)+sqrtx))=$

$=1+lim_(h->0) h/(h(sqrt(x+h)+sqrtx))=1+lim_(h->0) 1/(sqrt(x+h)+sqrtx)$

$=1+1/(sqrtx+sqrtx)=1+1/(2sqrtx)$

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