Derivative

[123]

Calculate the function \(\displaystyle f(x)=\sqrt{2x}+\frac{1}{x}\) the derivative at a point \(\displaystyle x=2\)

 

[Deleted member 4993]

Calculate the function \(\displaystyle f(x)=\sqrt{2x}+\frac{1}{x}\) the derivative at a point \(\displaystyle x=2\)

Exactly where are you stuck?

find derivative of ?(2x)

find derivative of 1/x

add those up

evaluate at x = 2

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[123]

so first i need find derivative of function and then add 2?

 

[Deleted member 4993]

so first i need find derivative of function and then add 2?

I am not sure what you mean by that?

 

[123]

what do with x=2?

 

[Deleted member 4993]

what do with x=2?

You evaluate the derivative at x = 2

 

[123]

\(\displaystyle f'(x)=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\)\(\displaystyle {\frac{2}{\sqrt{x}}\)???

 

[Deleted member 4993]

\(\displaystyle f'(x)=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\)\(\displaystyle {\frac{2}{\sqrt{x}}\)???

Incorrect

\(\displaystyle \frac{d(\sqrt{2x})}{dx} \ = \ \sqrt{2} \cdot \frac{1}{2} \frac{1}{\sqrt{x}}\)

now find derivative of (1/x)

 

[123]

\(\displaystyle {(\frac{1}{x})}'=-\frac{1}{x^2}\)?

 

[Deleted member 4993]

\(\displaystyle {(\frac{1}{x})}'=-\frac{1}{x^2}\)?

so

\(\displaystyle f(x) \ = \ \sqrt{2x} \ + \ \frac{1}{x}\)

\(\displaystyle \frac{df}{dx} \ = \ \frac{d( \sqrt{2x})}{dx} \ + \ \frac{d\left ( \frac{1}{x}\right )}{dx}\)

Now add those up to find f'(x) - then evaluate f'(2).

 

[123]

then answer is \(\displaystyle \frac{1}{4}\)

 

[Deleted member 4993]

then answer is \(\displaystyle \frac{1}{4}\)

How did you get that answer?