Finding Tangent Line to 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 h->0 which is the slope.

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

 

[HallsofIvy]

Do you mean that you have to use the basic formula for derivative
lim_{h->0}\frac{f(a+h)-f(a)}{h} rather than derivative formulas?

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

 

[Victor Frankenstein]

I did that and I got something like 1/2*Sqrt(a) by taking the lim as h->0, giving a slope of 1/4, but the answer is sopposed to be y=(1/20)*x+(1/5) can you please show me how they got this ?

By the way is that factor you posted called a conjugate ?