Question

ive never got the hang of first principles:
find from first principles the derivative of f(x) = x + sqrt(x).

Answer 1

\[f(x) = x + \sqrt{x}\]
\[f'(x) = (1)x ^{1-1} + (1/2) x^{\frac{1}{2} -1} \]
\[f'(x)= x^0 + \frac{1}{2} x^{-1/2}\]
\[f'(x)= 1 + \frac{1}{2 \sqrt{x}}\]

Answer 2

the question asks for first principles - as in the whole lim x as h->0 of f(x+h) - f(x)/h. thanks though

Answer 3

ready to do this?

Answer 4

\[f'(x) = {\lim_{h \to 0}}\; \; {f(x + h) - f(x) \over h} \]

Answer 5

Can you apply that here?

Answer 6

\[\lim_{h \to 0} \; \; {(x + h + \sqrt{x + h}) - (x + \sqrt{x}) \over h } \]\[\lim_{h \to 0 } \; \; {x + h + \sqrt{x + h} - x - \sqrt{x} \over h} \]

Answer 7

Apply L'Hospital's Rule to simplify the limit.

Answer 8

@ParthKohli No.

Answer 9

\[\lim_{h \rightarrow 0} \frac{ f(x+h)-f(x) }{ h }\] \[\lim_{h \rightarrow 0} \frac{ [(x+h)+\sqrt{x+h}] - (x+\sqrt{x}) }{ h }\] \[\lim_{h \rightarrow 0} \frac{ x+h+\sqrt{x+h}-x-\sqrt{x} }{ h }\] \[\lim_{h \rightarrow 0} \frac{ (1+\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x}) }{ h(\sqrt{x+h}+\sqrt{x}) }\] \[\lim_{h \rightarrow 0} \frac{ 1+(x+h-x) }{ h(\sqrt{x+h}+\sqrt{x}) }\] \[\lim_{h \rightarrow 0} 1+\frac{ h }{ h(\sqrt{x+h}+\sqrt{x}) }\] \[\lim_{h \rightarrow 0} 1+\frac{ 1 }{ \sqrt{x+h}+\sqrt{x} }\] Let h --> 0: \[1+\frac{ 1 }{ \sqrt{x+0}+\sqrt{x} }\] \[= 1+\frac{ 1 }{ 2\sqrt{x} }\]

Answer 10

OK, right. I actually might have forgotten that one, maybe. Doing the power rule has made me lazy ;)

Answer 11

@nj1202 mostly right 89% B+
read the Code of Conduct

Answer 12

I'm aware answer-giving is not allowed, but then I would've just put the last step :P

Answer 13

@nj1202 so stop doing entire problems for people. And if you are going to do entire problems, at least do them right.