Tangent Lines (2)Calculus
posted by Greg .
find an equation of the line tangent to y= sqrt(25x^2) at the point (3,4)
i got y= x/8 + 29/8
this is wrong. but im not sure why. can you explain thank you

y=√(25x^2)
y' = x/√(25x^2)
y'(3) = 3/4
so, now you have a point and a slope. The line is thus
y4 = 3/4 (x3)