Derivative at a Point problem

Question

Let p(x)=3 sqerroot x

1. Find the slop of the functions graph at x=25
2. Find an equation for the line tangent to the graph of y=p(x) at the point (x,y)=(25,15)

Answer 1

p(x) = 3sqr(x) = 3x^(1/2)

p'(x) = 3(1/2)x^(-1/2) (bring the power down as a coefficient and reduce the power by 1. 1/2 -1 = -1/2)

x^(-1/2) = 1/x^(1/2) = 1/sqr(x)

p'(x) = (3/2)/sqr(x)

p'(25) = (3/2)/sqr25 = (3/2)/5 = (3/2)(1/5) = 3/10 = .3 = the slope at x=25, at the point (25,15)

y-15 = (3/10)(x-25) is the equation of the tangent line in point slope form

or

10y -150 = 3x-75

3x-10y = -75 in standard form

or

y = (3/10)x -7.5+15

y = .3x +7.5 in slope intercept form

y = 3x/10 +15/2 with fractions