Finding Tangent Line of f(x) at (4,(2/5))

[Victor Frankenstein]

I need to find an equation for the tangent line to the graph of the function at the specified point, I had some trouble simplifing f(x) so that I can take the lim as x->0 which is the slope.

f(x) = Sqrt(x)/5 at (4,(2/5))

 

[bomba923]

*Simply find the derivative of the curve at that point, and use point-slope to find the line.
f\left( x \right) = \frac{\sqrt x}{5} \Rightarrow f\,'\left( x \right) = \frac{1}{{10\sqrt x }}

*To find the slope of the tangent line, simply calculate f\,' ( 4 ):
f\,'\left( 4 \right) = \frac{1}{{20}}

*Next, just use point-slope to represent the tangent line, which I'll call y:
\frac{{x - 4}}{{20}} = y - \frac{2}{5}

And it's algebra from there

 

[HallsofIvy]

I need to find an equation for the tangent line to the graph of the function at the specified point, I had some trouble simplifing f(x) so that I can take the lim as x->0 which is the slope.

f(x) = Sqrt(x)/5 at (4,(2/5))

Are you saying that you are required to use the basic formula for the derivative: lim_{h->0}\frac{f(a+h)-f(a)}{h} rather than the more specific formula (derived from that) that bomba923 used?

If so try multiplying both numerator and denominator of
\frac{\sqrt{a+h}-\sqrt{a}}{h}
by
\sqrt{a+h}+ \sqrt{a}.