Could somebody give me a hand with the following question:

Find an equation of the tangent line to the curve at the given point:

y=sqrt(x), (49,7)

Question

Could somebody give me a hand with the following question:

Find an equation of the tangent line to the curve at the given point:

y=sqrt(x), (49,7)

anonymous · Answer

$$
y'= \frac 1{2\sqrt x}
$$

anonymous · Answer

Find the slope m=y'(49) and us
y-y1 =m(x-x1)

anonymous · Answer

$$
y'(49)=\frac 1 { 2 \sqrt{49}}=\frac 1 {14}

$$

anonymous · Answer

$$

y-7 =\frac 1 {14} (x-49)
$$

anonymous · Answer

Thank you very much, I greatly appreciate your help, but I have another question to ask you if you don't mind.

anonymous · Answer

Since I'm just starting calculus, I have to take the long way around to get the solution; I have to use the equation for slope.  I don't suppose you could show me how that works could  you?

anonymous · Answer

Have they explained derivative to you?